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Fie \( n,k\in\mathbb{N}^{*} \) si \( a_{1},a_{2},\ldots,a_{n}\in\mathbb{R} \). Daca \( a_{1}^{2^{k}}+a_{2}^{2^{k}}+\ldots+a_{n}^{2^{k}}\le n^{2} \) sa se arate ca \( a_{1}+a_{2}+\ldots+a_{n}\le n^{2} \).

Din inegalitatea C.B.S.

\( (a_1^{2^{k-1}}+...+a_n^{2^{k-1}})^2\le n(a_1^{2^k}+...+a_n^{2^k})\le n^3\le n^4 \) de unde

\( a_1^{2^{k-1}}+...+a_n^{2^{k-1}}\le n^2 \)

Analog aplicand acest procedeu pas cu pas pana cand exponentii ajung la 1 ,obtinem in final concluzia dorita.