Inequalities & Convexity

Study Sheet

Inequalities & Convexity

AM–GM, Cauchy–Schwarz, Jensen, rearrangement, and integral inequalities

The Inequality Toolkit

Tip
The core four

AM–GM: ∑ain≥∏ain\tfrac{\sum a_i}{n} \ge \sqrt[n]{\prod a_i}. Cauchy–Schwarz: (∑aibi)2≤∑ai2∑bi2(\sum a_ib_i)^2 \le \sum a_i^2\sum b_i^2. Jensen: for convex ff, f(∑xin)≤∑f(xi)nf(\tfrac{\sum x_i}{n}) \le \tfrac{\sum f(x_i)}{n}. Rearrangement: similarly ordered sequences maximize ∑aibi\sum a_ib_i.

Each comes with an EQUALITY CASE — all equal for AM–GM, proportional sequences for Cauchy–Schwarz, all equal for strictly convex Jensen — and the equality case tells you which inequality to use: if the answer "should" be attained at a=b=ca = b = c, try AM–GM or Jensen. Cauchy–Schwarz in Engel form, ∑ai2bi≥(∑ai)2∑bi\sum\tfrac{a_i^2}{b_i} \ge \tfrac{(\sum a_i)^2}{\sum b_i}, proves (a+b+c)(1a+1b+1c)≥9(a+b+c)(\tfrac1a + \tfrac1b + \tfrac1c) \ge 9 in one line. Power means (QM≥AM≥GM≥HMQM \ge AM \ge GM \ge HM) are Jensen for xpx^p.

Tip
Convexity and its consequences

A function with f′′≥0f'' \ge 0 lies above its tangent lines and below its chords. Both facts are inequalities you can quote.

Above tangents: ex≥1+xe^x \ge 1 + x, ln⁡x≤x−1\ln x \le x - 1, sin⁡x≤x\sin x \le x for x≥0x \ge 0 (concave). Below chords: the average of ff at points is at least ff at the average (Jensen). Tangent-line trick for olympiad-style sums: bound each term f(xi)f(x_i) below by the tangent at the equality point, then sum — the constraint ∑xi=const\sum x_i = \text{const} makes the linear terms collapse. Integral versions: ∫01f≥f(12)\int_0^1 f \ge f(\tfrac12) for convex ff (Hermite–Hadamard), and (∫fg)2≤∫f2∫g2\left(\int fg\right)^2 \le \int f^2\int g^2 (Cauchy–Schwarz for integrals).

Side note
Weighted and integral forms are the same theorems

Cauchy–Schwarz is an inner-product statement, so it holds for sums, weighted sums, and integrals alike — (∫abfg)2≤∫abf2∫abg2\left(\int_a^bfg\right)^2 \le \int_a^bf^2\int_a^bg^2 — and Jensen holds for any probability weighting: f(E[X])≤E[f(X)]f(E[X]) \le E[f(X)]. A Putnam analysis problem that looks new is often Cauchy–Schwarz with an integral in place of a sum.

Try it
Try it: locate the equality case first

Minimize x+4xx + \dfrac4x for x>0x > 0, then minimize x2+2x2+1\dfrac{x^2 + 2}{\sqrt{x^2 + 1}} over all real xx.

AM–GM: x+4x≥24=4x + \tfrac4x \ge 2\sqrt4 = 4 at x=2x = 2. For the second, set t=x2+1≥1t = \sqrt{x^2+1} \ge 1: the expression is t+1t≥2t + \tfrac1t \ge 2 at t=1t = 1, i.e. x=0x = 0 — and t=1t = 1 IS in the domain, so the minimum 22 is attained.

Proofs & Why It Matters

Tip
Proof: Cauchy–Schwarz by a quadratic

For real vectors a,b\mathbf a, \mathbf b: (a⋅b)2≤∣a∣2∣b∣2(\mathbf a\cdot\mathbf b)^2 \le |\mathbf a|^2|\mathbf b|^2.

For every real tt, 0≤∣a−tb∣2=∣a∣2−2t(a⋅b)+t2∣b∣20 \le |\mathbf a - t\mathbf b|^2 = |\mathbf a|^2 - 2t(\mathbf a\cdot\mathbf b) + t^2|\mathbf b|^2. A quadratic in tt that is never negative has discriminant ≤0\le 0: 4(a⋅b)2−4∣a∣2∣b∣2≤04(\mathbf a\cdot\mathbf b)^2 - 4|\mathbf a|^2|\mathbf b|^2 \le 0. Equality iff the quadratic has a root, i.e. a=tb\mathbf a = t\mathbf b. ■\blacksquare The identical argument with ∫\int in place of ∑\sum proves the integral version — one proof, every inner product space.

Tip
Proof: Jensen's inequality

For convex ff and weights wi≥0w_i \ge 0 summing to 11: f(∑wixi)≤∑wif(xi)f(\sum w_ix_i) \le \sum w_if(x_i).

Let m=∑wixim = \sum w_ix_i and take a supporting line at mm: convexity gives f(x)≥f(m)+s(x−m)f(x) \ge f(m) + s(x - m) for all xx (with s=f′(m)s = f'(m) when differentiable). Apply at each xix_i, multiply by wiw_i, and add: ∑wif(xi)≥f(m)+s∑wi(xi−m)=f(m)+0\sum w_if(x_i) \ge f(m) + s\sum w_i(x_i - m) = f(m) + 0. ■\blacksquare Significance: AM–GM is Jensen for ln⁡\ln (concave), the power-mean inequality is Jensen for xpx^p, and entropy inequalities in information theory are Jensen for xln⁡xx\ln x — the tangent-line argument is the master proof.

Going Deeper: Worked Problems

Example
Worked: an integral inequality

For continuous f>0f > 0 on [0,1][0,1], show ∫01f⋅∫011f≥1\displaystyle\int_0^1 f\cdot\int_0^1\frac1f \ge 1.

Step 1 — Cauchy–Schwarz for integrals with f\sqrt f and 1f\tfrac{1}{\sqrt f}: (∫01f⋅1f)2≤∫01f⋅∫011f\left(\int_0^1\sqrt f\cdot\tfrac{1}{\sqrt f}\right)^2 \le \int_0^1 f\cdot\int_0^1\tfrac1f.

Step 2 — the left side is (∫011)2=1\left(\int_0^1 1\right)^2 = 1.

Step 3 — equality iff f\sqrt f and 1f\tfrac{1}{\sqrt f} are proportional, i.e. ff constant. This is the integral version of (∑ai)(∑1ai)≥n2(\sum a_i)(\sum\tfrac{1}{a_i}) \ge n^2 — the same inequality, the same proof, an integral in place of a sum.

Example
Worked: the tangent-line trick

For positive a,b,ca, b, c with a+b+c=3a + b + c = 3, show ∑aa2+1≤32\displaystyle\sum\frac{a}{a^2 + 1} \le \frac32.

Step 1 — the equality case is a=b=c=1a = b = c = 1, where each term is 12\tfrac12.

Step 2 — find the tangent line to g(x)=xx2+1g(x) = \tfrac{x}{x^2+1} at x=1x = 1: g′(x)=1−x2(x2+1)2g'(x) = \tfrac{1 - x^2}{(x^2+1)^2}, so g′(1)=0g'(1) = 0 and the tangent is the horizontal line y=12y = \tfrac12.

Step 3 — claim g(x)≤12g(x) \le \tfrac12 for all x>0x > 0: indeed xx2+1≤12  ⟺  2x≤x2+1  ⟺  (x−1)2≥0\tfrac{x}{x^2+1} \le \tfrac12 \iff 2x \le x^2 + 1 \iff (x-1)^2 \ge 0 ✓.

Step 4 — sum: ∑g(a)≤32\sum g(a) \le \tfrac32. Here the constraint was not even needed — but in general the tangent line has a slope ss, and ∑s(xi−1)=s(∑xi−3)=0\sum s(x_i - 1) = s(\sum x_i - 3) = 0 is where the constraint enters.