What does Jensen's inequality say about a convex function of a weighted average?

## Statement Let `f` be convex on a convex domain `D`, let `x_1, ..., x_n` be points of `D`, and let `t_1, ..., t_n` be weights with ``` t_i >= 0 and t_1 + ... + t_n = 1 ``` Then **Jensen's inequality** says ``` f( t_1*x_1 + ... + t_n*x_n ) <= t_1*f(x_1) + ... + t_n*f(x_n) ``` In words: **the function of the weighted average is at most the weighted average of the function values**. If `f` is concave, every inequality reverses. If `f` is affine, both directions hold and the relation is an equality. The hypotheses carry real weight. The `x_i` must lie in a region where `f` is convex — a formula convex on one interval need not be convex on a larger one. The weights must be non-negative and sum to one; a negative weight breaks the argument immediately, because multiplying an inequality by a negative number flips it. ## Why it is true The two-point case `f(t*x_1 + (1-t)*x_2) <= t*f(x_1) + (1-t)*f(x_2)` is precisely the definition of a convex function, so there is nothing to prove there. For `n` points, argue by induction. Assume the result for `n - 1` points and suppose `t_n < 1` (if `t_n = 1` the statement is trivial). Write ``` sum_{i=1..n} t_i*x_i = (1 - t_n) * z + t_n * x_n, where z = sum_{i<n} [t_i / (1 - t_n)] * x_i ``` The coefficients `t_i / (1 - t_n)` for `i < n` are non-negative and sum to `1`, so `z` is a convex combination of the first `n - 1` points and therefore lies in `D`. Apply the two-point inequality to `z` and `x_n`, then the induction hypothesis to `f(z)`, and the terms reassemble into the claim. Geometrically the picture is the same one as the chord test: the weighted average of the points `(x_i, f(x_i))` on the graph is a point inside their convex hull, which sits on or above the graph, and its height is `sum t_i f(x_i)` while the graph directly below it has height `f(sum t_i x_i)`. ## Equality For an affine `f`, Jensen is an identity for every choice of points and weights. For a **strictly** convex `f`, equality holds if and only if all the `x_i` that carry positive weight are equal to one another — any genuine spread among the weighted points makes the inequality strict. This equality case is the source of most of the sharpness results proved with Jensen: you get a bound, and then the equality condition tells you exactly which configuration attains it. ## Deriving AM-GM The classic deterministic payoff. The natural logarithm is **concave** on the positive reals (its second derivative is `-1/x^2 < 0`), so Jensen applies with the reversed inequality. Take positive numbers `x_1, ..., x_n` and equal weights `t_i = 1/n`: ``` log( (x_1 + ... + x_n) / n ) >= (1/n) * ( log x_1 + ... + log x_n ) = log( (x_1 * ... * x_n)^(1/n) ) ``` Since `log` is strictly increasing, exponentiating preserves the direction: ``` (x_1 + ... + x_n) / n >= (x_1 * x_2 * ... * x_n)^(1/n) ``` which is exactly **AM-GM**: the arithmetic mean is at least the geometric mean. Because `log` is strictly concave, equality holds if and only if all the `x_i` coincide. The same argument with general weights `t_i` gives the **weighted AM-GM**: ``` t_1*x_1 + ... + t_n*x_n >= x_1^{t_1} * ... * x_n^{t_n} ``` ## How it is used Jensen is the standard tool for turning a statement about a single averaged quantity into a statement about individual terms, or the reverse. Whenever a bound involves a convex function on the outside of a sum, Jensen lets you push the function inside the sum at a known cost — and knowing which direction the cost points is the whole skill. The most common error in interviews is stating the inequality with the sign reversed, so anchor it on a concrete case you can verify in your head: with `f(x) = x^2` and the two numbers `0` and `2` with equal weights, `f(1) = 1` while the average of the function values is `2`, and `1 <= 2` confirms the direction.