Inequalities & Convexity

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 $a = b = c$, try AM–GM or Jensen. Cauchy–Schwarz in Engel form, $\sum\tfrac{a_i^2}{b_i} \ge \tfrac{(\sum a_i)^2}{\sum b_i}$, proves $(a+b+c)(\tfrac1a + \tfrac1b + \tfrac1c) \ge 9$ in one line. Power means ($QM \ge AM \ge GM \ge HM$) are Jensen for $x^p$.

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

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

Let $m = \sum w_ix_i$ and take a supporting line at $m$: convexity gives $f(x) \ge f(m) + s(x - m)$ for all $x$ (with $s = f'(m)$ when differentiable). Apply at each $x_i$, multiply by $w_i$, and add: $\sum w_if(x_i) \ge f(m) + s\sum w_i(x_i - m) = f(m) + 0$. $\blacksquare$ Significance: AM–GM is Jensen for $\ln$ (concave), the power-mean inequality is Jensen for $x^p$, and entropy inequalities in information theory are Jensen for $x\ln x$ — the tangent-line argument is the master proof.

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

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

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

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

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

Step 3 — claim $g(x) \le \tfrac12$ for all $x > 0$: indeed $\tfrac{x}{x^2+1} \le \tfrac12 \iff 2x \le x^2 + 1 \iff (x-1)^2 \ge 0$ ✓.

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