Olympiad MethodsDifficulty 65-1001 tagged problems

Jensen's Inequality

Jensen's Inequality is the habit of applying convexity or concavity to bound averages of function values. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

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The Key Result

$$\text{For convex } f:\quad f\!\left(\frac{x_1 + \cdots + x_n}{n}\right) \;\le\; \frac{f(x_1) + \cdots + f(x_n)}{n}.$$

For a convex function, the value at the average of inputs is at most the average of the values (the inequality reverses for concave $f$); equality holds when all inputs are equal or $f$ is linear on the relevant range.

Why It Works

1
A function $f$ is convex on an interval if for all $x,y$ in it and $t\in[0,1]$, $f(tx + (1-t)y) \le t f(x) + (1-t) f(y)$ — the chord lies above the graph.
2
Apply this two-point definition repeatedly (or by induction on $n$) to the weighted average; for equal weights $\frac{1}{n}$ it generalizes to $f\!\left(\frac{1}{n}\sum x_i\right) \le \frac{1}{n}\sum f(x_i)$.
3
More generally, for weights $w_i \ge 0$ with $\sum w_i = 1$, convexity gives $f\!\left(\sum w_i x_i\right) \le \sum w_i f(x_i)$ — the full weighted Jensen inequality.
4
If $f$ is concave (e.g. $\log$, $\sqrt{\cdot}$), the inequality flips: $f\!\left(\frac{\sum x_i}{n}\right) \ge \frac{\sum f(x_i)}{n}$. Equality requires all $x_i$ equal (when $f$ is strictly convex).

Worked Examples

Example 1

For angles $A, B, C$ of a triangle, prove $\sin A + \sin B + \sin C \le \frac{3\sqrt{3}}{2}$.

1
On $(0,\pi)$ the function $\sin x$ has second derivative $-\sin x < 0$, so it is concave there; the angles satisfy $A+B+C = \pi$ with each in $(0,\pi)$.
2
Apply Jensen for a concave function: $\frac{\sin A + \sin B + \sin C}{3} \le \sin\!\left(\frac{A+B+C}{3}\right)$.
3
The argument is $\frac{A+B+C}{3} = \frac{\pi}{3}$, where $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$.
4
Multiply by $3$: $\sin A + \sin B + \sin C \le 3\cdot\frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$, with equality at the equilateral triangle $A=B=C=\frac{\pi}{3}$.

Answer:

$\sin A + \sin B + \sin C \le \dfrac{3\sqrt{3}}{2}$, equality at $A=B=C=\dfrac{\pi}{3}$.

Contest level

For positive reals $x_1, \dots, x_n$, prove the power-mean inequality $\dfrac{x_1^2 + \cdots + x_n^2}{n} \ge \left(\dfrac{x_1 + \cdots + x_n}{n}\right)^2$.

1
The function $f(t) = t^2$ has $f''(t) = 2 > 0$, so it is convex on all of $\mathbb{R}$.
2
Jensen for convex $f$ with equal weights: $f\!\left(\tfrac{1}{n}\sum x_i\right) \le \tfrac{1}{n}\sum f(x_i)$.
3
Substitute $f(t) = t^2$: $\left(\tfrac{1}{n}\sum x_i\right)^2 \le \tfrac{1}{n}\sum x_i^2$, which is exactly the claim.
4
Equality holds iff all $x_i$ are equal (since $t^2$ is strictly convex).

Answer:

$\dfrac{\sum x_i^2}{n} \ge \left(\dfrac{\sum x_i}{n}\right)^2$, equality iff all $x_i$ equal — the QM $\ge$ AM inequality.

Olympiad / Challenge

For the angles $A, B, C$ of a triangle, prove $\tan\dfrac{A}{2} + \tan\dfrac{B}{2} + \tan\dfrac{C}{2} \ge \sqrt{3}$.

1
The half-angles satisfy $\tfrac{A}{2} + \tfrac{B}{2} + \tfrac{C}{2} = \tfrac{\pi}{2}$, with each in $\left(0, \tfrac{\pi}{2}\right)$.
2
On $\left(0, \tfrac{\pi}{2}\right)$, $f(x) = \tan x$ is convex: $f''(x) = 2\sec^2 x\,\tan x > 0$ there.
3
Jensen for a convex function gives $\tfrac{1}{3}\sum \tan\tfrac{A}{2} \ge \tan\!\left(\tfrac{1}{3}\cdot\tfrac{\pi}{2}\right) = \tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt3}$.
4
Multiply by $3$: $\sum \tan\tfrac{A}{2} \ge \tfrac{3}{\sqrt3} = \sqrt3$, with equality at the equilateral triangle $A = B = C = \tfrac{\pi}{3}$.

Answer:

$\tan\tfrac{A}{2} + \tan\tfrac{B}{2} + \tan\tfrac{C}{2} \ge \sqrt3$, equality at the equilateral triangle.

Going Deeper

Weighting Jensen with $w_i = \tfrac{1}{n}$ and $f = \log$ (concave) recovers AM–GM; with general weights it yields the weighted AM–GM and the full power-mean hierarchy $M_{-\infty} \le \cdots \le M_0 \le \cdots \le M_\infty$.
Jensen is the most-used single inequality on olympiad inequality problems; the 'tangent line trick' (bounding $f$ below by its tangent at the equality point) is a hand-computable substitute when the constraint isn't a plain average.
Pitfall: convexity must point the RIGHT way over the WHOLE relevant interval. $\sin$ is concave on $(0,\pi)$ (so the average-of-values is an UPPER bound), while $\tan$ is convex on $(0,\tfrac\pi2)$; applying Jensen with the wrong sign — or where $f$ changes concavity — gives a false inequality.

Spot the Signal

Look for problems where the key step is applying convexity or concavity to bound averages of function values.
You can describe the hard part as applying convexity or concavity to bound averages of function values, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for jensen's inequality, then rewrite the givens around it.
Name the relation that makes Jensen's Inequality legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Jensen's Inequality just because the surface looks familiar; verify the required condition first.
Applying Jensen's Inequality because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let jensen's inequality reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is applying convexity or concavity to bound averages of function values.