ExampleConstruction: AI-adaptedVerification: AI-generatedaudited 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 $λ_{i}$ with $\sum_{i} λ_{i} = 1$ and real $x_{i}$ ,

${(\sum_{i} λ_{i} x_{i})}^{2} \leq \sum_{i} λ_{i} x_{i}^{2} .$

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

$λ x^{2} + (1 - λ) y^{2} - (λ x + (1 - λ) y)^{2} = λ (1 - λ) (x - y)^{2} \geq 0$

for $0 \leq λ \leq 1$ , 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↦x2 gives that the square of a weighted mean is at most the weighted mean of the squares