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id: inequalities-olympiad title: Advanced Inequalities author: USAMO Guide Team — Aayush Bathija description: Techniques in homogenization, substitution, and inequality proofs. prerequisites: ["inequalities-advanced", "proof-writing-basics"]

Overview

Inequalities at the olympiad level demand structured reasoning rather than brute-force computation. A central skill is selecting transformations and substitutions that allow the application of standard inequalities in a clean and effective way.

Key Ideas

* Homogenize expressions to ensure all terms have the same degree. * Use substitutions such as a=xya = xy, b=yzb = yz, c=zxc = zx or a=xy,;b=yz,;c=zxa = \frac{x}{y},; b = \frac{y}{z},; c = \frac{z}{x} in symmetric settings. * Identify equality cases early to guide the direction of the proof.

Core Skills

Homogenization

In many inequalities, given conditions can make direct approaches cumbersome. In such cases, homogenization is a key technique.

Example

Given abc=1, prove that a2+b2+c2≥a+b+c.\text{Given } abc = 1,\ \text{prove that } a^2 + b^2 + c^2 \ge a + b + c.

At first glance, this inequality may seem difficult to approach. A useful observation is that most classical inequalities (such as AM-GM or Cauchy-Schwarz) are naturally applied to homogeneous expressions.

An expression is called homogeneous if all its terms have the same total degree. This property is powerful because it allows scaling: variables can be multiplied by a constant without changing the structure of the inequality.

To homogenize, rewrite the inequality using the condition abc=1abc = 1:

a2+b2+c2≥a(abc)13+b(abc)13+c(abc)13.a^2 + b^2 + c^2 \ge a(abc)^{\frac{1}{3}} + b(abc)^{\frac{1}{3}} + c(abc)^{\frac{1}{3}}.

Now both sides have degree 2, and the inequality follows immediately from AM-GM:

a2≥a,b2≥b,c2≥c.a^2 \ge a,\quad b^2 \ge b,\quad c^2 \ge c.

Summing yields the desired result. □\square

Substitution

While homogenization is effective, substitutions can often streamline the process.

Common substitutions include:

* If abc=1abc = 1, let

a=xy,b=yz,c=zx.a = \frac{x}{y},\quad b = \frac{y}{z},\quad c = \frac{z}{x}.

* For expressions involving reciprocals, consider

a=1x,b=1y,c=1z.a = \frac{1}{x},\quad b = \frac{1}{y},\quad c = \frac{1}{z}.

* In symmetric expressions, substitutions such as

a=xy,b=yz,c=zxa = xy,\quad b = yz,\quad c = zx

can simplify the structure and reveal hidden symmetry.

Choosing the right substitution is often guided by experience and by examining the form of the inequality.

Proof Techniques

To prove an inequality of the form A≥CA \ge C, a common strategy is to construct a chain:

A≥B≥D≥C.A \ge B \ge D \ge C.

Finding appropriate intermediate expressions is often the central challenge.

Another key technique is analyzing equality cases:

* If equality occurs when a=b=ca = b = c, this strongly suggests using symmetric inequalities such as AM-GM. * If equality depends on proportional relationships, inequalities like Cauchy-Schwarz may be more appropriate.

It is often helpful to determine equality conditions at the outset, as they provide direction for the proof.

Common Pitfalls

* Ignoring or failing to check equality cases * Reversing inequality directions during transformations * Overlooking domain constraints (e.g., positivity conditions)

Practice Problems

StatusSourceProblem NameDifficultyTags
Harvard-MIT Mathematics TournamentHard
Show TagsCatalan numbers, partitions, Muirhead / majorization
Team Selection Test 2009Hard
Show TagsMuirhead / majorization, QM-AM-GM-HM / Power Mean, Symmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsIntermediate Value Theorem, Muirhead / majorization, Polynomial operations, Symmetric functions, Vieta's formulas
Berkeley Math CircleHard
Show TagsMuirhead / majorization, QM-AM-GM-HM / Power Mean, Symmetric functions
Berkeley Math CircleHard
Show TagsColoring schemes, extremal arguments, Combinatorial optimization, Jensen / smoothing
Berkeley Math Circle Monthly Contest 3Hard
Show TagsInvariants / monovariants, Jensen / smoothing, Sums and products
BAMOHard
Show TagsGraph Theory, Jensen / smoothing, Linear and quadratic inequalities
Berkeley Math Circle Monthly Contest #1Hard
Show TagsColoring schemes, extremal arguments, Combinatorial optimization, Jensen / smoothing, Sums and products
USA IMOHard
Show TagsJensen / smoothing, Linear and quadratic inequalities, QM-AM-GM-HM / Power Mean, Quadratic functions, Triangle trigonometry, Trigonometry
Harvard-MIT November TournamentHard
Show TagsJensen / smoothing, QM-AM-GM-HM / Power Mean
2nd Annual Harvard-MIT November TournamentHard
Show TagsAlgorithms, Invariants / monovariants, Jensen / smoothing
Bay Area Mathematical OlympiadHard
Show TagsAngle chasing, Cauchy-Schwarz, Jensen / smoothing, Jensen/smoothing, QM-AM-GM-HM / Power Mean, Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle, Triangle trigonometry, Trigonometry
Team Selection Test Selection TestHard
Show TagsJensen / smoothing, Jensen/smoothing, Rotation, Tangents, Translation, Triangle inequalities, Triangle trigonometry, Trigonometry, Vectors
Berkeley Math Circle Monthly Contest 4Hard
Show TagsCombinatorial optimization, Jensen / smoothing, QM-AM-GM-HM / Power Mean
HMMT 2014Hard
Show TagsJensen / smoothing
Berkeley Math Circle: Monthly Contest 2Hard
Show TagsExpected values, Jensen / smoothing
Berkeley Math Circle: Monthly Contest 5Hard
Show TagsColoring schemes, extremal arguments, Counting two ways, Jensen / smoothing
Berkeley Math CircleHard
Show TagsInduction / smoothing, Jensen / smoothing, Linear and quadratic inequalities
HMMT NovemberHard
Show TagsAlgebraic properties of binomial coefficients, Jensen / smoothing, Volume
HMMT NovemberHard
Show TagsCombinatorial optimization, Jensen / smoothing
February 2017Hard
Show TagsCombinatorial optimization, Jensen / smoothing
HMMT November 2019Hard
Show TagsJensen / smoothing
HMMT Spring 2021 Guts RoundHard
Show TagsFactorization techniques, Induction / smoothing, Invariants / monovariants, Jensen / smoothing, Sums and products

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