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Daca \( a_1,\ a_2,\ \dots ,\ a_n > 0 \) cu \( a_1^{2}+a_2^{2}+\dots+a_n^{2}=1 \), atunci:

\( a_1+a_2+\dots+a_n+\frac{1}{a_1a_2... a_n}\ge \sqrt{n}+\sqrt{n^n} \)

\( a_1+a_2+...+a_n+\frac{1}{a_1a_2...a_n}=(a_1+a_2+...+a_n)(a_1^2+a_2^2+...+a_n^2)^{\frac{n-1}{2}}+\frac{1}{a_1a_2...a_n}\ge n\sqrt[n]{a_1a_2...a_n}(n\sqrt[n]{a_1a_2...a_n})^{\frac{n-1}{2}}+\frac{1}{a_1a_2...a_n}=n^{\frac{n-1}{2}}p+\frac{1}{p},\ \mbox{unde}\ p=a_1a_2...a_n \)

Astfel ramane de aratat ca \( n^{\frac{n-1}{2}}p+\frac{1}{p}\ge \sqrt{n}+\sqrt{n^n} \)

\( \Longleftrightarrow (1-p\sqrt{n})(1-p\sqrt{n^n})\ge 0 \), inegalitate care este adevarata deoarece avem \( 1=a_1^2+a_2^2+...+a_n^2\ge n\sqrt[n]{a_1^2a_2^2...a_n^2}=np^{\frac{2}{n}}. \). Egalitatea are loc pentu \( a_1=a_2=...=a_n=\frac{1}{\sqrt{n^n}}. \)