ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)audited 2026-08-02

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Finite Jensen for $x\mapsto x^2$ gives that the square of a weighted mean is at most the weighted mean of the squares

Example

For nonnegative weights $\lambda_i$ with $\sum_i\lambda_i=1$ and real $x_i$ ,

$\left(\sum_i\lambda_i x_i\right)^2\le\sum_i\lambda_i x_i^2.$

The function $q(t)=t^2$ is convex because

$\lambda x^2+(1-\lambda)y^2-(\lambda x+(1-\lambda)y)^2=\lambda(1-\lambda)(x-y)^2\ge0$

for $0\le\lambda\le1$ , using nonnegativity of squares (Squares of nonzero elements are positive). Applying finite Jensen (Finite Jensen inequality for a convex function and nonnegative weights summing to one) to $q$ gives the displayed inequality.

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Sources

S. Boyd and L. Vandenberghe, Convex Optimization, §3.1 (standard reference, not scraped)

Finite Jensen for x↦x2x\mapsto x^2x↦x2 gives that the square of a weighted mean is at most the weighted mean of the squares