Part 4: Advanced Inequalities and Majorization

The full algebraic and analytic arsenal — majorization theory, Schur convexity, Muirhead's theorem, the SOS algorithm, Vasc's EV method, L^p integral bounds, and the Rearrangement inequality — deployed against the deadliest competition problems.

Prerequisite: Mastery, Not Familiarity

This handout is not an introduction. It is written for those who have already conquered the standard curriculum — you can differentiate in your sleep, convexity is second nature, and AM-GM is a reflex. If you’re here, you’ve cleared the floor that 99% never reach. What follows is the architecture above it: the theorems, algorithms, and structural insights that separate the top solvers from everyone who merely “knows the material.” Proceed accordingly.

In Part 1 , we laid the algebraic and topological foundations. In Part 2 , we forged the differential tools — recursive derivatives, the wavy curve, and sign analysis. In Part 3 , we deployed the inequality machinery — the Mean Value Theorems, convexity, and a six-technique heuristic arsenal. Now we ascend to the full landscape.

This is where the top 1% separate from the merely excellent. The tools in this module — Schur convexity, Muirhead’s theorem, the Sum of Squares algorithm, Vasc’s Equal Variables method, the L^p integral bounds (Hölder, Minkowski, Young), and the Rearrangement inequality — are the weapons that obliterate problems which resist every classical approach. Each framework here has destroyed entire generations of Olympiad contestants who lacked the vocabulary to even name the technique they needed.

I. The Majorization Deepening

In Part 3, Section II , we defined majorization and stated Karamata’s inequality. Here we excavate the structural foundations — the matrix-theoretic characterization that reveals why majorization works, and the differential criterion that makes Schur convexity computable.

The Hardy-Littlewood-Pólya Theorem

Hardy-Littlewood-Pólya — The Matrix Characterization

The vector $\mathbf{x}$ majorizes $\mathbf{y}$ (i.e., $\mathbf{x} \succ \mathbf{y}$ ) if and only if there exists a doubly stochastic matrix $P$ such that $\mathbf{y} = P\mathbf{x}$ .

A doubly stochastic matrix is a square matrix of non-negative reals where every row and every column sums to exactly $1$ .

Why this matters: The theorem says that $\mathbf{y}$ being majorized by $\mathbf{x}$ is equivalent to $\mathbf{y}$ being a “weighted average” of the components of $\mathbf{x}$ . The vector $\mathbf{y}$ is literally a smoothed version of $\mathbf{x}$ .

Birkhoff’s Theorem and the Geometric Picture

Birkhoff's Theorem

The set of all $n \times n$ doubly stochastic matrices forms a convex polytope whose extreme points (vertices) are precisely the $n!$ permutation matrices.

Combining Hardy-Littlewood-Pólya with Birkhoff: $\mathbf{y}$ is majorized by $\mathbf{x}$ if and only if $\mathbf{y}$ can be written as a convex combination of permutations of $\mathbf{x}$ . In competition terms, if you can show one vector is a convex mix of permutations of another, majorization is immediate.

Schur Convexity

A symmetric function $F: \mathbb{R}^n \to \mathbb{R}$ is Schur-convex if $\mathbf{x} \succ \mathbf{y}$ implies $F(\mathbf{x}) \ge F(\mathbf{y})$ .

Equivalently, $F$ is Schur-convex if it preserves the majorization ordering.

Schur-Ostrowski Criterion — The Differential Test

A continuously differentiable symmetric function $F$ is Schur-convex if and only if for all pairs $i \neq j$ :

$(x_i - x_j)\!\Bigl(\frac{\partial F}{\partial x_i} - \frac{\partial F}{\partial x_j}\Bigr) \ge 0$

Deploying the Schur-Ostrowski Criterion

To prove $F(\mathbf{x}) \ge F(\mathbf{y})$ where $\mathbf{x} \succ \mathbf{y}$ :

Verify symmetry: Confirm $F$ is symmetric under variable permutation.
Compute partials: Find $\frac{\partial F}{\partial x_i}$ and $\frac{\partial F}{\partial x_j}$ .
Factor: Compute the difference $\frac{\partial F}{\partial x_i} - \frac{\partial F}{\partial x_j}$ and factor out $(x_i - x_j)$ .
Sign check: If the remaining factor is globally non-negative, $F$ is Schur-convex.

The collapse: Once Schur-convexity is established, the minimum of $F$ under a fixed sum constraint $\sum x_i = S$ occurs at the completely averaged vector $(S/n, \ldots, S/n)$ , and the maximum at the maximally skewed vector $(S, 0, \ldots, 0)$ .

II. Symmetric Polynomial Bounds

Muirhead’s Theorem

Muirhead's Inequality

Let $\boldsymbol{\alpha} = (\alpha_1, \ldots, \alpha_n)$ and $\boldsymbol{\beta} = (\beta_1, \ldots, \beta_n)$ be exponent sequences, and define the symmetric mean:

$[\boldsymbol{\alpha}] = \frac{1}{n!} \sum_{\sigma \in S_n} x_{\sigma(1)}^{\alpha_1} x_{\sigma(2)}^{\alpha_2} \cdots x_{\sigma(n)}^{\alpha_n}$

For all positive reals $x_1, \ldots, x_n$ : $[\boldsymbol{\alpha}] \ge [\boldsymbol{\beta}]$ if and only if $\boldsymbol{\alpha} \succ \boldsymbol{\beta}$ .

Exponent $\boldsymbol{\alpha}$	Exponent $\boldsymbol{\beta}$	Majorization Check	Result
$(5, 0, 0)$	$(3, 1, 1)$	$5 \ge 3$ , $5 \ge 4$ , $5 = 5$ ✓	$\sum_{\text{sym}} x^5 \ge \sum_{\text{sym}} x^3yz$
$(4, 2, 0)$	$(3, 3, 0)$	$4 \ge 3$ , $6 \ge 6$ , $6 = 6$ ✓	$\sum_{\text{sym}} x^4y^2 \ge \sum_{\text{sym}} x^3y^3$
$(3, 2, 1)$	$(4, 1, 1)$	$3 \not\ge 4$ ✗	Reversed — bound goes the other way

Muirhead as Oracle, AM-GM as Proof

In stringent grading environments (IMO, Putnam written), citing “by Muirhead” is strategically risky — it can be perceived as invoking a black box. Muirhead’s true competition value is as a heuristic oracle:

Check feasibility: Does $\boldsymbol{\alpha} \succ \boldsymbol{\beta}$ ? If not, the bound is impossible.
Construct the proof: Build an explicit weighted AM-GM chain that evaluates to the same symmetric result.
Submit the AM-GM proof: This is universally accepted without further justification.

The oracle tells you what is true. AM-GM proves why.

Trap 1: The Cyclic Sum Trap

Muirhead’s theorem requires the full symmetric sum (averaging over all $n!$ permutations). A cyclic sum over three variables like $\sum_{\text{cyc}} a^3b = a^3b + b^3c + c^3a$ involves only 3 terms, not the $6$ terms of the symmetric sum $\sum_{\text{sym}} a^3b$ .

Applying majorization logic to cyclic permutations produces mathematically false results. This error on an Olympiad paper results in immediate penalization.

The Sum of Squares (SOS) Method

SOS Canonical Form

For a symmetric inequality $F(a, b, c) \ge 0$ with three variables, the SOS decomposition writes:

$F(a, b, c) = S_a(b - c)^2 + S_b(c - a)^2 + S_c(a - b)^2$

Since $(b - c)^2, (c - a)^2, (a - b)^2 \ge 0$ , proving $F \ge 0$ reduces to analyzing the coefficient signs $S_a, S_b, S_c$ .

Assuming WLOG $a \ge b \ge c$ , the algebraic identity $(a - c)^2 \ge (a - b)^2 + (b - c)^2$ (since $2(a - b)(b - c) \ge 0$ ) yields the following positivity criteria:

Known Signs	Required Criterion
$S_a, S_b, S_c \ge 0$	None — trivially non-negative
$S_a \ge 0$ , $S_b \ge 0$ , $S_c < 0$	$S_b + S_c \ge 0$
$S_c \ge 0$ , $S_b \ge 0$ , $S_a < 0$	$S_a + S_b \ge 0$
$S_a < 0$ and $S_c < 0$	$a^2 S_b + b^2 S_a \ge 0$ and $b^2 S_c + c^2 S_b \ge 0$

The SOS Algorithm

Expand $F(a, b, c)$ and collect terms.
Extract squared differences: use the identity $(a - b)(a - c) = \frac{1}{2}((a - b)^2 + (a - c)^2 - (b - c)^2)$ and similar.
Collect coefficients of $(b - c)^2$ , $(c - a)^2$ , $(a - b)^2$ across all cyclic shifts.
Check the SOS positivity criteria under the ordering $a \ge b \ge c$ .

Power: SOS handles asymmetric fractional denominators and radical roots by systematically multiplying out conjugates to extract $(x - y)^2$ factors.

III. Calculus-Driven Optimization

Vasc’s Equal Variables (EV) Method

While SOS handles algebraic reductions, Vasile Cîrtoaje’s Equal Variables method provides a calculus-driven topological approach. It resolves optimization bounds by proving that extremal values of symmetric functions under sum or product constraints occur when at least two variables are equal.

Cîrtoaje's Half-Convex Function Theorem

Let $f: I \to \mathbb{R}$ have exactly one inflection point $s \in I$ , with $f$ convex on $\{u \ge s\}$ and concave on $\{u \le s\}$ . Then the extrema of $\sum_{i=1}^n f(x_i)$ subject to $\sum x_i = \text{constant}$ occur when at least $n - 1$ of the variables are equal.

The EV Reduction Pipeline

For a symmetric inequality $F(a, b, c) \ge 0$ with $a + b + c = S$ :

Identify the component function: Write $F = \sum f(x_i)$ or relate $F$ to such a sum.
Find the inflection point: Compute $f''(x)$ and locate the sign change.
Apply the Half-Convex theorem: The minimum occurs at $(a, t, t)$ or $(t, t, c)$ for some values.
Substitute: Set $b = c = t$ , use $a = S - 2t$ , and reduce to a single-variable polynomial.
Factor: The resulting polynomial typically factors as $(t - t_0)^2 \cdot Q(t)$ with $Q > 0$ .

Why this is devastating: It systematically obliterates symmetric inequalities up to degree 8 that completely resist AM-GM, Schur, or Jensen.

Trap 2: Unsorted Majorization

When applying Karamata’s inequality, sequences must be sorted in descending order before checking partial sum conditions. If $x_1 = 1, x_2 = 5$ and $y_1 = 3, y_2 = 3$ , checking $x_1 \ge y_1$ yields $1 \ge 3$ — false. Sort first to $\mathbf{x} = (5, 1)$ , $\mathbf{y} = (3, 3)$ , then $5 \ge 3$ and $6 = 6$ . ✓

IV. L^p Integral Bounds

The discrete algebraic frameworks above govern the finite sequences of Olympiad algebra. The continuous counterparts — governing integrals over measure spaces — dominate the subjective calculus sections of ISI B.Math, CMI entrance, and advanced Putnam problems. Three inequalities form the complete hierarchy.

Young’s Inequality

Young's Inequality

Let $p, q > 1$ satisfy the conjugate relation $\frac{1}{p} + \frac{1}{q} = 1$ . For any non-negative reals $a, b \ge 0$ :

$ab \le \frac{a^p}{p} + \frac{b^q}{q}$

The geometric proof: Consider a strictly increasing function $f(x) = x^{p-1}$ with inverse $f^{-1}(y) = y^{q-1}$ . The rectangle with vertices at the origin, $(a, 0)$ , $(a, b)$ , $(0, b)$ has area $ab$ . This area is bounded by the sum of the area under $f$ from $0$ to $a$ and the area to the left of $f$ from $0$ to $b$ :

$ab \le \int_0^a x^{p-1}\, dx + \int_0^b y^{q-1}\, dy = \frac{a^p}{p} + \frac{b^q}{q}$

This visceral geometric argument — bounding a rectangle by two complementary curved regions — is the deepest way to understand why products are bounded by scaled powers.

Hölder’s Inequality

Hölder's Inequality

For measurable functions $f, g$ and conjugate exponents $p, q$ with $\frac{1}{p} + \frac{1}{q} = 1$ :

$\int_X |f(x) g(x)|\, d\mu \le \left(\int_X |f(x)|^p\, d\mu\right)^{1/p} \left(\int_X |g(x)|^q\, d\mu\right)^{1/q}$

Special case: $p = q = 2$ recovers the Cauchy-Schwarz inequality for integrals.

The Normalization Technique (Hölder's Proof Engine)

The standard proof of Hölder’s inequality uses integral normalization — a meta-technique that appears repeatedly in ISI/CMI subjective problems:

Define normalized functions: $F(x) = \frac{|f(x)|}{(\int |f|^p)^{1/p}}$ and $G(x) = \frac{|g(x)|}{(\int |g|^q)^{1/q}}$ .
This scaling forces $\int F^p = 1$ and $\int G^q = 1$ .
Apply Young’s pointwise: $F(x)G(x) \le \frac{1}{p}F(x)^p + \frac{1}{q}G(x)^q$ .
Integrate: $\int FG \le \frac{1}{p} + \frac{1}{q} = 1$ .
Unpack the normalization to recover the standard Hölder form.

When to deploy: Any integral problem where you need to bound $\int fg$ by separate norms of $f$ and $g$ .

Minkowski’s Inequality

Minkowski's Inequality

For $p \ge 1$ and measurable functions $f, g$ :

$\left(\int_X |f + g|^p\, d\mu\right)^{1/p} \le \left(\int_X |f|^p\, d\mu\right)^{1/p} + \left(\int_X |g|^p\, d\mu\right)^{1/p}$

This establishes that the L^p norm satisfies the triangle inequality, making $L^p$ a Banach space.

The derivation cascade: Factor $|f+g|^p = |f+g| \cdot |f+g|^{p-1}$ , bound via triangle inequality, apply Hölder to each resulting integral, factor out $(\int |f+g|^p)^{1/q}$ , and divide. The exponent arithmetic $1 - 1/q = 1/p$ closes the proof.

Framework	Domain	Core Statement	Primary Use
Young’s	Pointwise	$ab \le \frac{a^p}{p} + \frac{b^q}{q}$	Converting products to additive bounds
Hölder’s	Integral	$\int \\|fg\\| \le \\|f\\|_p \\|g\\|_q$	Bounding integrated products (ISI subjective)
Minkowski’s	Integral	$\\|f+g\\|_p \le \\|f\\|_p + \\|g\\|_p$	Triangle inequality for functional interpolation

V. Permutation Optimization

The Rearrangement Inequality

Given sorted sequences $a_1 \le a_2 \le \cdots \le a_n$ and $b_1 \le b_2 \le \cdots \le b_n$ , and any permutation $\sigma \in S_n$ :

$\sum_{i=1}^n a_i b_{n+1-i} \le \sum_{i=1}^n a_i b_{\sigma(i)} \le \sum_{i=1}^n a_i b_i$

The dot product is maximized when the sequences are similarly sorted and minimized when oppositely sorted.

The swapping proof: Assume a permutation $\sigma$ achieves the maximum but is not the identity. Then there exist $i < j$ with $\sigma(i) > \sigma(j)$ . Since $a_i \le a_j$ and $b_{\sigma(i)} \ge b_{\sigma(j)}$ , swapping the pairings changes the sum by $(a_j - a_i)(b_{\sigma(i)} - b_{\sigma(j)}) \ge 0$ . Repeating until no misaligned pairs remain forces the identity permutation.

Chebyshev’s Sum Inequality

Chebyshev's Sum Inequality

If $(a_i)$ and $(b_i)$ are similarly sorted, then:

$n \sum_{i=1}^n a_i b_i \ge \left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n b_i\right)$

For oppositely sorted sequences, the inequality reverses.

Derivation from Rearrangement: Consider the $n$ cyclic shifts of $(b_i)$ . By Rearrangement, the identity permutation sum $\sum a_i b_i$ dominates each cyclic shift. Summing all $n$ inequalities yields $n \sum a_i b_i \ge (\sum a_i)(\sum b_i)$ .

Alignment Paradigms

Before applying Rearrangement or Chebyshev:

Explicitly establish sorting: Prove $a_1 \le a_2 \le \cdots \le a_n$ and determine the sorting of $(b_i)$ .
Watch for entanglement: If the rank ordering of $(a_i)$ depends on a parameter that simultaneously alters the ordering of $(b_i)$ , the inequality cannot be directly invoked.
Construct auxiliary sequences: When entanglement exists, define $c_i = f(a_i)$ to artificially force monotonic alignment before applying the inequality.

Grading requirement: In written Olympiads, establishing the explicit ordering proof is mandatory. Skipping it produces a “fake-solve.”

Trap 3: Alignment Entanglement

Error: Applying the Rearrangement inequality to sequences whose monotonic ordering depends on an algebraic parameter that simultaneously affects both sequences.

If the sorting of $(a_i)$ changes with a parameter $t$ that also changes the sorting of $(b_i)$ , the inequality’s hypothesis is violated. Advanced Putnam problems deliberately exploit this trap.

Fix: Construct auxiliary sequences with provably fixed orderings before invoking the inequality.

VI. Advanced Manipulation Heuristics

Three algebraic micro-heuristics allow the major theorems above to bypass strict structural constraints.

1. Isolated Fudging (The Tangent Line Trick Without Convexity)

Building on the tangent line method from Part 3, Section III , isolated fudging extends the technique to problems where global convexity fails:

Identify the equality case: For a constrained sum $\sum x_i = S$ , equality typically occurs at $x_i = S/n = c$ for all $i$ .
Compute the tangent line: $T(x) = f'(c)(x - c) + f(c)$ .
Prove the algebraic bound $f(x) \ge T(x)$ directly — by factoring the difference polynomial. Often $(x - c)^2$ divides the difference.
Sum: $\sum f(x_i) \ge \sum T(x_i) = nf(c)$ since $\sum (x_i - c) = 0$ .

The key difference from Part 3: This works even when $f''$ changes sign — you bypass convexity entirely by proving the tangent bound algebraically via polynomial factorization.

2. Rational Homogenization

Inequalities conditioned on a constant product (e.g., $abc = 1$ ) often prevent polynomial degree matching. Homogenize via:

$a = \frac{x}{y}, \quad b = \frac{y}{z}, \quad c = \frac{z}{x}$

This ensures $abc = 1$ unconditionally for all $x, y, z > 0$ , freeing the inequality for attack by Muirhead’s or Schur’s theorems without the product constraint interfering.

When to use: Any time a product constraint prevents direct degree comparison or AM-GM alignment.

3. Trigonometric Substitution for Algebraic Constraints

For structural conditions like $xy + yz + zx = 1$ , the substitution $x = \tan(A/2)$ , $y = \tan(B/2)$ , $z = \tan(C/2)$ — where $A, B, C$ are angles of a triangle — ensures the constraint is inherently satisfied.

This transforms rigid algebraic bounds into trigonometric optimization, opening the door for Jensen’s inequality on concave functions like $\sin$ and $\cos$ .

Signal: When you see $xy + yz + zx$ appearing as a constraint alongside expressions involving $\sqrt{1 + x^2}$ , the tangent half-angle substitution is almost certainly the intended path.

Trap 4: Homogenization Degree Mismatch

Error: Attempting to apply Muirhead’s theorem to an inequality whose left and right sides have different polynomial degrees.

Muirhead compares symmetric sums of the same total degree. If the LHS has degree 7 and the RHS has degree 5, you must homogenize by multiplying the lower-degree side by $(xyz)^{k/3}$ (using the constraint) to equalize degrees before applying the theorem.

Fix: Always verify degree matching before invoking Muirhead. Use the constraint to embed the necessary powers.

Trap 5: Black Box Citation

Error: Writing “by Muirhead’s theorem” as the sole justification in a written Olympiad proof.

In stringent grading environments (IMO, Putnam), this can be penalized as invoking an unproven result. Muirhead’s theorem is not on the “universally accepted without proof” list (unlike AM-GM).

Fix: Use Muirhead as a heuristic oracle to identify the correct bound, then construct an explicit AM-GM chain that proves it.

VII. Beyond the Syllabus

Operator Inequalities and Matrix Convexity

The theory of majorization extends far beyond real vectors. For Hermitian matrices $A$ and $B$ , the vector of eigenvalues of $A + B$ is majorized by the sum of eigenvalue vectors of $A$ and $B$ . This is proved via the Rayleigh quotient characterization of eigenvalues and Birkhoff’s polytope machinery. Lieb’s Concavity Theorem — that certain trace-exponential functions of Hermitian matrices preserve concavity — is the frontier, connecting the convexity hierarchy of Part 3 to quantum information theory.

Schur Polynomials and Representation Theory

The symmetric means $[\boldsymbol{\alpha}]$ from Muirhead’s theorem are special cases of Schur polynomials — the characters of irreducible representations of the general linear group $GL(n)$ . The majorization ordering on exponent vectors corresponds precisely to the dominance ordering on partitions, which governs the decomposition of tensor products of representations. When a competition problem asks you to compare symmetric polynomial sums, you are — perhaps unknowingly — navigating the lattice of Young diagrams.

Concentration Inequalities and Measure Theory

In probability theory, the L^p bounds (Hölder, Minkowski) are the backbone of concentration inequalities — results that show random variables cluster near their expectations. Markov’s inequality ( $P(X \ge a) \le E[X]/a$ ) is a direct consequence of Hölder with $p = 1$ , $q = \infty$ . The Chernoff bound, which governs tail probabilities of independent sums, uses the exponential moment technique: apply Hölder to $e^{tX}$ . These tools are the bridge between the calculus inequalities of this module and the probabilistic world of information theory, statistical mechanics, and machine learning.

Illustrative Examples

Problem Hard

(Isolated Fudging / Tangent Line Trick)

Let $a, b, c, d > 0$ with $a + b + c + d = 1$ . Prove that $6(a^3 + b^3 + c^3 + d^3) \ge a^2 + b^2 + c^2 + d^2 + \dfrac{1}{8}$ .

View Solution

Step 1 — Identify the equality case: By symmetry and the constraint, equality should occur at $a = b = c = d = \frac{1}{4}$ .

Step 2 — Define the component function: Let $f(x) = 6x^3 - x^2$ . The inequality becomes $\sum f(a_i) \ge \frac{1}{8}$ .

Step 3 — Compute the tangent at the equality point: $f(1/4) = 6/64 - 1/16 = 1/32$ . $f'(x) = 18x^2 - 2x$ , so $f'(1/4) = 18/16 - 1/2 = 5/8$ .

The tangent line at $x = 1/4$ : $T(x) = \frac{5x - 1}{8}$ .

Step 4 — Prove $f(x) \ge T(x)$ algebraically: The difference is:

$6x^3 - x^2 - \frac{5x - 1}{8} = \frac{48x^3 - 8x^2 - 5x + 1}{8}$

Since the tangent touches at $x = 1/4$ , the factor $(x - 1/4)^2 = (4x - 1)^2/16$ must divide the numerator. By polynomial division:

$48x^3 - 8x^2 - 5x + 1 = 3(4x - 1)^2\!\left(x + \frac{1}{3}\right)$

For $x > 0$ : $(4x - 1)^2 \ge 0$ and $(x + 1/3) > 0$ , so the product is non-negative. ✓

Step 5 — Sum over the variables:

$\sum f(a_i) \ge \sum T(a_i) = \sum \frac{5a_i - 1}{8} = \frac{5(a + b + c + d) - 4}{8} = \frac{5 - 4}{8} = \frac{1}{8}$

Equality if and only if $a = b = c = d = 1/4$ . $\blacksquare$

15 min USA Math Olympiad — Tangent Line Paradigm

Problem Advanced

(The SOS Algorithm)

For positive reals $a, b, c > 0$ , prove that $\displaystyle\sum_{\text{cyc}} \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} \ge 0$ .

View Solution

Step 1 — Extract SOS structure: Use the identity $(a-b)(a-c) = \frac{1}{2}((a-b)^2 + (a-c)^2 - (b-c)^2)$ .

Substituting into the cyclic sum and collecting coefficients of $(b-c)^2$ across all three permutations yields the canonical SOS form $\sum_{\text{cyc}} S_a(b - c)^2 \ge 0$ .

Step 2 — Compute $S_a$ : Tracking the $(b-c)^2$ coefficient across cyclic shifts:

$S_a = \frac{1}{2b^2 + (c+a)^2} + \frac{1}{2c^2 + (a+b)^2} - \frac{1}{2a^2 + (b+c)^2}$

Step 3 — Sign analysis under $a \ge b \ge c$ : This ordering implies:

$2a^2 + (b+c)^2 \ge 2b^2 + (c+a)^2 \ge 2c^2 + (a+b)^2$

Denominators shrink, so reciprocals grow. Therefore $\frac{1}{2c^2 + (a+b)^2} \ge \frac{1}{2a^2 + (b+c)^2}$ , guaranteeing $S_a \ge 0$ . By identical reasoning, $S_b \ge 0$ .

Step 4 — Handle the potentially negative $S_c$ : The SOS criterion requires $S_b + S_c \ge 0$ . Computing:

$S_b + S_c = \frac{1}{2c^2 + (a+b)^2} + \frac{1}{2a^2 + (b+c)^2} - \frac{1}{2b^2 + (c+a)^2} + \frac{1}{2a^2 + (b+c)^2} + \frac{1}{2b^2 + (c+a)^2} - \frac{1}{2c^2 + (a+b)^2}$

After cancellation: $S_b + S_c = \frac{2}{2a^2 + (b+c)^2} > 0$ since $a, b, c > 0$ .

Step 5 — Conclude: All SOS positivity criteria are satisfied. Therefore $\sum_{\text{cyc}} \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} \ge 0$ . Equality if and only if $a = b = c$ . $\blacksquare$

18 min ELMO Shortlist 2010, adapted

Problem Advanced

(Vasc’s Equal Variables Method)

Let $a, b, c \ge 0$ with $a + b + c = 3$ . Prove that $3(a^4 + b^4 + c^4) + a^2 + b^2 + c^2 + 6 \ge 6(a^3 + b^3 + c^3)$ .

View Solution

Step 1 — Setup: Define $F(a, b, c) = 3(a^4 + b^4 + c^4) - 6(a^3 + b^3 + c^3) + (a^2 + b^2 + c^2) + 6$ . We need $F \ge 0$ .

Step 2 — Apply the mixing variables displacement: WLOG $a \le b \le c$ . Let $t = \frac{b+c}{2}$ and $x = \frac{c - b}{2}$ , so $b = t - x$ and $c = t + x$ . Compute $\Delta = F(a, b, c) - F(a, t, t)$ .

Using the algebraic expansions:

$b^2 + c^2 - 2t^2 = 2x^2$
$b^3 + c^3 - 2t^3 = 6tx^2$
$b^4 + c^4 - 2t^4 = 12t^2 x^2 + 2x^4$

Step 3 — Compute the displacement:

$\Delta = 3(12t^2 x^2 + 2x^4) - 6(6tx^2) + 2x^2 = 2x^2(18t^2 - 18t + 1 + 3x^2)$

Step 4 — Reduce to equal variables: By the Half-Convex Function Theorem, it suffices to prove $F(a, t, t) \ge 0$ where $a + 2t = 3$ , i.e., $a = 3 - 2t$ . Substituting:

$F(3 - 2t, t, t) = 3((3-2t)^4 + 2t^4) - 6((3-2t)^3 + 2t^3) + ((3-2t)^2 + 2t^2) + 6$

Step 5 — Factor the single-variable polynomial: Expanding and simplifying:

$F(3 - 2t, t, t) = 54(t - 1)^2(t^2 - t + 1)$

Since $(t - 1)^2 \ge 0$ always, and $t^2 - t + 1 = (t - 1/2)^2 + 3/4 > 0$ has no real roots, the product is strictly non-negative.

Step 6 — Conclude: The global minimum occurs at $a = b = c = 1$ , where $F(1, 1, 1) = 0$ . Therefore $F(a, b, c) \ge 0$ unconditionally. $\blacksquare$

20 min Adapted from V. Cîrtoaje, Mathematical Inequalities Vol. 1

Problem Hard

(Muirhead’s Theorem with Homogenization)

For positive reals $x, y, z$ with $xyz = 1$ , prove that $\dfrac{x^3}{y^2} + \dfrac{y^3}{z^2} + \dfrac{z^3}{x^2} \ge x^2 + y^2 + z^2$ .

Direct Muirhead on the cyclic sum

The LHS $\sum_{\text{cyc}} x^3/y^2$ is a cyclic sum, not a symmetric sum. Muirhead’s theorem requires averaging over all $n! = 6$ permutations, not just the 3 cyclic ones. Attempting to apply Muirhead directly here is the Cyclic Sum Trap (Trap 1) — the result would be mathematically invalid.

View Solution

Step 1 — Eliminate denominators: Since $xyz = 1$ , we have $x^2 y^2 z^2 = 1$ , so $x^2 y^2 = 1/z^2$ . Therefore:

$\frac{x^3}{y^2} = \frac{x^5}{x^2 y^2} = x^5 z^2$

The inequality transforms to $\sum_{\text{cyc}} x^5 z^2 \ge \sum_{\text{cyc}} x^2$ .

Step 2 — Homogenize to equal degree: The LHS has degree $5 + 2 = 7$ . The RHS has degree 2. Multiply the RHS by $(xyz)^{5/3} = 1$ to match: $x^2 \cdot (xyz)^{5/3} = x^{11/3} y^{5/3} z^{5/3}$ .

Step 3 — Symmetrize the LHS: The symmetric sum $\sum_{\text{sym}} x^5 y^0 z^2$ has exponent vector $\boldsymbol{\alpha} = (5, 2, 0)$ and the target has $\boldsymbol{\beta} = (11/3, 5/3, 5/3)$ .

Step 4 — Verify majorization:

$5 \ge 11/3$ ✓
$5 + 2 = 7 \ge 11/3 + 5/3 = 16/3$ ✓
$5 + 2 + 0 = 7 = 11/3 + 5/3 + 5/3$ ✓

Since $(5, 2, 0) \succ (11/3, 5/3, 5/3)$ , Muirhead’s theorem gives $[\boldsymbol{\alpha}] \ge [\boldsymbol{\beta}]$ . (Note: Muirhead’s theorem extends to real — not just integer — exponents for positive variables, as the symmetric mean $[\boldsymbol{\alpha}]$ is well-defined for all $\alpha_i \in \mathbb{R}$ when $x_i > 0$ .)

Step 5 — Construct the AM-GM chain: For the constructive proof, by weighted AM-GM:

$\frac{10}{21}x^5 z^2 + \frac{1}{21}y^5 x^2 + \frac{10}{21}z^5 y^2 \ge x^{11/3} y^{5/3} z^{5/3}$

Summing cyclically over all three variables completes the proof. $\blacksquare$

15 min IMO Shortlist — Homogenization Paradigm

Problem Hard

(Continuous Inequality via Monotonicity and Integration)

Let $f: [1, \infty) \to \mathbb{R}$ be differentiable with $f(1) = 1$ and $f'(x) = \dfrac{1}{x^2 + f(x)^2}$ . Prove that $f(x) \le 1 + \dfrac{\pi}{4}$ for all $x \ge 1$ .

View Solution

Step 1 — Establish monotonicity: Since $x^2 \ge 1$ on $[1, \infty)$ and $f^2(x) \ge 0$ , the denominator $x^2 + f^2(x) > 0$ . Thus $f'(x) > 0$ — the function is strictly increasing.

Step 2 — Bound $f$ from below: Since $f$ is increasing and $f(1) = 1$ , we have $f(x) \ge 1$ for all $x \ge 1$ . Therefore $f^2(x) \ge 1$ .

Step 3 — Bound $f'$ from above: Adding $x^2$ to both sides of $f^2(x) \ge 1$ :

$x^2 + f^2(x) \ge x^2 + 1 \implies f'(x) = \frac{1}{x^2 + f^2(x)} \le \frac{1}{x^2 + 1}$

Step 4 — Integrate: For any $x \ge 1$ :

$f(x) - f(1) = \int_1^x f'(t)\, dt \le \int_1^x \frac{dt}{t^2 + 1} = \arctan(x) - \arctan(1)$

Step 5 — Bound the arctangent: Since $\arctan$ is bounded above by $\pi/2$ :

$\arctan(x) - \arctan(1) < \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

Therefore $f(x) \le 1 + \frac{\pi}{4}$ for all $x \ge 1$ . $\blacksquare$

12 min ISI TOMATO Subjective 144

Problem Hard

(AM-GM on Geometric Series Terms)

For $x > 0$ and positive integer $n \ge 1$ , prove that $\dfrac{x^n - 1}{x - 1} \ge n\, x^{(n-1)/2}$ .

View Solution

Step 1 — Recognize the sum structure: The LHS is the geometric series sum:

$\frac{x^n - 1}{x - 1} = x^{n-1} + x^{n-2} + \cdots + x + 1$

Step 2 — Apply AM-GM: The arithmetic mean of the $n$ terms $x^0, x^1, \ldots, x^{n-1}$ is bounded below by their geometric mean:

$\frac{x^{n-1} + x^{n-2} + \cdots + 1}{n} \ge (x^0 \cdot x^1 \cdot x^2 \cdots x^{n-1})^{1/n}$

Step 3 — Compute the geometric mean: The exponent sum is $0 + 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2}$ . Thus:

$\text{GM} = \bigl(x^{n(n-1)/2}\bigr)^{1/n} = x^{(n-1)/2}$

Step 4 — Conclude: Multiplying both sides by $n$ :

$x^{n-1} + x^{n-2} + \cdots + 1 \ge n\, x^{(n-1)/2}$

Equality holds if and only if all terms are equal, i.e., $x = 1$ (verified via L’Hôpital: $\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n = n \cdot 1^{(n-1)/2}$ ). $\blacksquare$

10 min ISI TOMATO Subjective 77

Problem Advanced

(Rearrangement Inequality with Trigonometric Substitution)

Let $x, y, z > 0$ with $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} > x + y + z$ . Prove that $\dfrac{2x}{\sqrt{1 + x^2}} + \dfrac{2y}{\sqrt{1 + y^2}} + \dfrac{2z}{\sqrt{1 + z^2}} \le 3$ .

View Solution

Step 1 — Decode the constraint: Since $x, y, z > 0$ , for each variable $\frac{1}{x} > x$ implies $x < 1$ (as $1 > x^2$ for positive $x$ ). More precisely, multiplying the original inequality by $xyz > 0$ gives $yz + xz + xy > xyz(x + y + z)$ , but the key structural observation is that by AM-GM on each pair $\frac{1}{x_i} - x_i > 0$ , all variables satisfy $0 < x_i < 1$ . The tangent half-angle substitution below handles the constraint algebraically: the condition $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} > x + y + z$ for positive reals maps to $A + B + C < \pi$ .

Step 2 — Trigonometric substitution: Let $x = \tan(A/2)$ , $y = \tan(B/2)$ , $z = \tan(C/2)$ where $A, B, C > 0$ . The constraint $xy + yz + zx < 1$ corresponds to $A + B + C < \pi$ — the angles form an acute triangle.

Step 3 — Simplify the target expression: For each term:

$\frac{x}{\sqrt{1 + x^2}} = \frac{\tan(A/2)}{\sec(A/2)} = \sin(A/2)$

The inequality becomes $2\sin(A/2) + 2\sin(B/2) + 2\sin(C/2) \le 3$ .

Step 4 — Apply Jensen’s Inequality: Define $g(\theta) = 2\sin(\theta/2)$ . Then $g''(\theta) = -\frac{1}{2}\sin(\theta/2) < 0$ on $(0, \pi)$ , so $g$ is strictly concave.

By Jensen for concave functions:

$g(A) + g(B) + g(C) \le 3\, g\!\Bigl(\frac{A + B + C}{3}\Bigr)$

Since $A + B + C < \pi$ and $g$ is increasing on $(0, \pi)$ :

$3\, g\!\Bigl(\frac{A + B + C}{3}\Bigr) < 3\, g\!\Bigl(\frac{\pi}{3}\Bigr) = 3 \cdot 2\sin\!\Bigl(\frac{\pi}{6}\Bigr) = 3 \cdot 2 \cdot \frac{1}{2} = 3 \quad \blacksquare$

18 min Singapore Mathematical Olympiad

Problem Advanced

(Deriving Minkowski’s Inequality from Hölder’s)

Using Hölder’s inequality as the sole base bounding mechanism, rigorously derive Minkowski’s inequality: $\left(\int_X |f+g|^p\, d\mu \right)^{1/p} \le \left(\int_X |f|^p\, d\mu\right)^{1/p} + \left(\int_X |g|^p\, d\mu\right)^{1/p}$ for $p > 1$ .

View Solution

Step 1 — Factor the exponent: Write $|f+g|^p = |f+g| \cdot |f+g|^{p-1}$ . By the triangle inequality, $|f+g| \le |f| + |g|$ , so:

$\int_X |f+g|^p\, d\mu \le \int_X |f| \cdot |f+g|^{p-1}\, d\mu + \int_X |g| \cdot |f+g|^{p-1}\, d\mu$

Step 2 — Apply Hölder to each term: Let $q$ be the Hölder conjugate of $p$ , i.e., $\frac{1}{p} + \frac{1}{q} = 1$ , which gives $(p-1)q = p$ . Applying Hölder’s inequality to the first integral:

$\int_X |f| \cdot |f+g|^{p-1}\, d\mu \le \left(\int_X |f|^p\, d\mu\right)^{1/p} \left(\int_X |f+g|^{(p-1)q}\, d\mu\right)^{1/q} = \|f\|_p \cdot \|f+g\|_p^{p/q}$

By identical logic for the $g$ term: $\int_X |g| \cdot |f+g|^{p-1}\, d\mu \le \|g\|_p \cdot \|f+g\|_p^{p/q}$ .

Step 3 — Combine and simplify: Adding both bounds:

$\|f+g\|_p^p \le \bigl(\|f\|_p + \|g\|_p\bigr) \cdot \|f+g\|_p^{p/q}$

Step 4 — Divide by the common factor: Dividing both sides by $\|f+g\|_p^{p/q}$ (valid when non-zero):

$\|f+g\|_p^{p - p/q} \le \|f\|_p + \|g\|_p$

Since $p - p/q = p \cdot (1 - 1/q) = p \cdot \frac{1}{p} = 1$ , the left side reduces to $\|f+g\|_p$ . Therefore:

$\|f+g\|_p \le \|f\|_p + \|g\|_p \quad \blacksquare$

20 min Functional Analysis — Hölder's Factorization

Selected Problems

Problem Hard

Let $x_1, \ldots, x_n$ be positive reals with $\prod_{i=1}^n x_i = 1$ . Prove that $\displaystyle\sum_{i=1}^n \frac{1}{1 + x_i} \le \sum_{i=1}^n \frac{x_i^k}{1 + x_i}$ for any $k > 1$ .

Hint

Assume WLOG that the array

x_i

is sorted in increasing order. Prove that the sequences

(x_i^k)

and

\bigl(\frac{1}{1+x_i}\bigr)

are oppositely sorted. Then apply Chebyshev’s sum inequality.

Problem Hard

For positive reals $a, b, c$ , prove that $(a^2 + b^2 + c^2)^2 \ge 3(a^3 b + b^3 c + c^3 a)$ .

Hint

This is a cyclic, non-symmetric inequality — Muirhead cannot be applied directly. Use the SOS algorithm: rewrite the polynomial difference

F(a,b,c)

into the canonical quadratic form

\sum S_a(b-c)^2

and verify the positivity criteria under

a \ge b \ge c

Problem Advanced

Let $P(z) = \sum_{k=0}^{2019} b_k z^k$ be a polynomial with complex roots $z_1, z_2, \ldots, z_{2019}$ . If $1 \le b_0 < b_1 < \cdots < b_{2019}$ , find the minimum possible average distance of the roots to the origin, i.e., minimize $\frac{1}{2019}\sum_{i=1}^{2019} |z_i|$ .

Hint

Relate the complex roots to the sequence coefficients via Vieta’s symmetric formulas. Apply Hölder’s discrete inequality to bound the absolute value sum

\sum |z_i|

. The monotone coefficient condition constrains the root structure.

Problem Advanced

Let $a, b, c > 0$ with $abc = 1$ . Prove that $\dfrac{1}{a^3(b+c)} + \dfrac{1}{b^3(c+a)} + \dfrac{1}{c^3(a+b)} \ge \dfrac{3}{2}$ .

Hint

Execute rational homogenization by substituting

a = 1/x

b = 1/y

c = 1/z

. The constraint becomes

xyz = 1

and the LHS transforms into

\sum \frac{x^2 y z}{y + z}

. Apply the Cauchy-Schwarz inequality in Engel’s fractional form (Titu’s Lemma).

Problem Hard

Prove Karamata’s inequality for $n = 3$ explicitly: if $f$ is convex and $(x_1, x_2, x_3) \succ (y_1, y_2, y_3)$ , then $f(x_1) + f(x_2) + f(x_3) \ge f(y_1) + f(y_2) + f(y_3)$ .

Hint

For each

y_i

, construct a secant line of

f

passing through two appropriate points among

(x_j, f(x_j))

. Use the convexity of

f

to bound

f(y_i)

above by the secant value. Sum, and exploit the majorization conditions

\sum x_i = \sum y_i

and the partial sum inequalities to cancel the linear terms.

Problem Hard

For non-negative reals $a, b, c$ , prove that $(a^2 + 2bc)(b^2 + 2ca)(c^2 + 2ab) \ge (ab + bc + ca)^3$ .

Hint

Deploy the Vasc EV Method. By symmetry, the minimum over the constraint manifold occurs when two variables are equal. Set

a = b = t

and reduce to a single-variable inequality in

t

and

c

. Factor the resulting polynomial to confirm non-negativity.

Problem Advanced

Suppose $A, B$ are positive definite $n \times n$ matrices. Prove that the vector of eigenvalues of $A + B$ is majorized by the sum of the eigenvalue vectors of $A$ and $B$ .

Hint

Use the Rayleigh quotient characterization of eigenvalues:

\lambda_k(M) = \max_{\dim V = k} \min_{x \in V, \|x\|=1} x^T M x

. The subadditivity of the Rayleigh quotient gives the partial sum inequalities. Connect to Birkhoff’s polytope to establish the doubly stochastic matrix relationship.

Problem Advanced

Let $f$ be continuous and non-negative on $[a, b]$ . Find the sharp upper bound for $\dfrac{\bigl(\int_a^b f(x)\, dx\bigr)^3}{\int_a^b f(x)^3\, dx}$ .

Hint

Employ Hölder’s inequality for integrals with conjugate pair

p = 3

q = 3/2

acting on

f(x)

and the constant function

1

. The sharp bound is

(b - a)^2

, achieved when

f

is constant. Show this by normalizing the integral and applying Young’s inequality.

Challenge Problem

Problem Advanced

(The Synthesis Problem)

Let $a, b, c$ be strictly positive real numbers. Prove that:

$\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \ge 1$

Hint 1: The Isolated Fudging Motivation

The inequality is homogeneous of degree zero. Hypothesize an algebraic bound of the form

\frac{a}{\sqrt{a^2 + 8bc}} \ge \frac{a^r}{a^r + b^r + c^r}

, which would automatically sum to

1

. To find the optimal exponent

r

, equate the partial derivatives of both sides with respect to

a

at the symmetry point

(1, 1, 1)

. The left derivative evaluates to

\frac{8}{27}

; the right to

\frac{2r}{9}

. Equating gives

r = \frac{4}{3}

Hint 2: The Convexity Transformation

Since the inequality is homogeneous, substitute

x = \frac{bc}{a^2}

y = \frac{ca}{b^2}

z = \frac{ab}{c^2}

to obtain

xyz = 1

and the cleaner form:

\frac{1}{\sqrt{1 + 8x}} + \frac{1}{\sqrt{1 + 8y}} + \frac{1}{\sqrt{1 + 8z}} \ge 1

. Let

x = e^U

y = e^V

z = e^W

U + V + W = 0

. Study the convexity of

g(T) = \frac{1}{\sqrt{1 + 8e^T}}

Hint 3: The EV Reduction

The function

g(T)

has exactly one inflection point and transitions from concave to convex — so Jensen’s inequality cannot be applied globally. Instead, apply Cîrtoaje’s Half-Convex Function Theorem: the minimum of

\sum g(T_i)

subject to

\sum T_i = 0

occurs when two variables are equal. Setting

y = z

forces

x = 1/y^2

, reducing the entire problem to a single-variable inequality that can be verified algebraically.