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Exercise 1731Algebra › InequalitiesL1Added 3 Nov 2026

The weighted AM-GM inequality

#weighted AM-GM#inequality#convexity

Statement

Let a1,…,an>0a_1,\ldots,a_n>0 and λ1,…,λn>0\lambda_1,\ldots,\lambda_n>0 with ∑λi=1\sum\lambda_i=1. Prove that
∏i=1naiλi≤∑i=1nλiai.\prod_{i=1}^n a_i^{\lambda_i}\le\sum_{i=1}^n\lambda_i a_i.

Hint

Use the concavity of ln⁡\ln and Jensen's inequality.

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Solution

Logarithm. Since ln⁡\ln is concave, Jensen's inequality gives
ln⁡(∑iλiai)≥∑iλiln⁡(ai)=ln⁡(∏iaiλi).\ln\left(\sum_i\lambda_i a_i\right)\ge\sum_i\lambda_i\ln(a_i)=\ln\left(\prod_i a_i^{\lambda_i}\right).

Exponentiation. Since exp⁡\exp is increasing:
∑iλiai≥∏iaiλi.\sum_i\lambda_i a_i\ge\prod_i a_i^{\lambda_i}.

Equality case. Equality holds iff all aia_i are equal (the concavity is strict).

Check. With λi=1/n\lambda_i=1/n, we recover the classical AM-GM.