Pieņemsim, ka \(x_{1},\ x_{2},\ \ldots,\ x_{n}\) ir nenegatīvi reāli skaitļi, \(n \geq 2\). Noskaidrot, kurām \(n\) vērtībām nevienādība
\[\frac{\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{2}^{2}+x_{3}^{2}\right) \ldots\left(x_{n-1}^{2}+x_{n}^{2}\right)\left(x_{n}^{2}+x_{1}^{2}\right)}{2^{n}} \geq\left(\frac{x_{1}x_{2}+x_{2}x_{3}+\ldots+x_{n-1}x_{n}+x_{n}x_{1}}{n}\right)^{n}\]
ir identiski patiesa.
Pie \(n=2\) nevienādība ir \(\frac{\left(x_{1}^{2}+x_{2}^{2}\right)^{2}}{4} \geq\left(\frac{x_{1} x_{2}+x_{2}x_{1}}{2}\right)^{2}\), kas reducējas par \(\left(x_{1}-x_{2}\right)^{2} \geq 0\) un ir identiski patiesa.
Pie \(n \geq 4\) nevienādība ir aplama, ja \(x_{1}=x_{2}=0\) un \(x_{3}=x_{4}=\ldots=x_{n}=1\).
Apskatām \(n=3\). Apzīmējam \(S_{1}=x_{1}+x_{2}+x_{3},\ S_{2}=x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1},\ S_{3}=x_{1}x_{2}x_{3}\). Ievēroja, ka nevienādības pareizība vai nepareizība nemainās, ja visus \(x_{i}\) dala ar vienu un to pašu pozitīvu skaitli. Izdarām to tā, lai būtu \(S_{2}=1\). (To nevar izdarīt, ja vismaz divi no \(x_{i}\) ir \(0\), bet tad nevienādība ir pareiza.) Tad mūsu nevienādība kļūst par
\[\frac{1}{8}\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{2}^{2}+x_{3}^{2}\right)\left(x_{3}^{2}+x_{1}^{2}\right) \geq \frac{1}{27}\]
Ievērojam, ka \(S_{1}^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2S_{2}\). Tā kā \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \geq S_{2}\) (tas seko no \(\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2} \geq 0\)), tad \(S_{1}^{2} \geq 3\) un \(S_{1} \geq \sqrt{3}\). No nevienādības starp vidējo aritmētisko un vidējo ģeometrisko iegūstam \(\frac{1}{3}=\frac{S_{2}}{3} \geq \sqrt[3]{S_{3}^{2}}\), tāpēc \(S_{3} \leq \frac{1}{3 \sqrt{3}}\) Tāpēc\[\begin{aligned} & \frac{1}{8}\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{2}^{2}+x_{3}^{2}\right)\left(x_{3}^{2}+x_{1}^{2}\right) \geq\left(\frac{x_{1}+x_{2}}{2}\right)^{2} \cdot\left(\frac{x_{2}+x_{3}}{2}\right)^{2} \cdot\left(\frac{x_{3}+x_{1}}{2}\right)^{2}= \\ & =\frac{1}{64}\left(S_{1}-x_{3}\right)^{2}\left(S_{1}-x_{1}\right)^{2}\left(S_{1}-x_{2}\right)^{2}=\frac{1}{64}\left[\left(S_{1}-x_{1}\right)\left(S_{1}-x_{2}\right)\left(S_{1}-x_{3}\right)\right]^{2}= \\ & =\frac{1}{64}\left[S_{1}^{3}-S_{1}^{2}\left(x_{1}+x_{2}+x_{3}\right)+S_{1} \cdot S_{2}-S_{3}\right]^{2}=\frac{1}{64}\left(S_{1}^{3}-S_{1}^{3}+S_{1}S_{2}-S_{3}\right)^{2}= \\ & =\frac{1}{64}\left(S_{1}S_{2}-S_{3}\right)^{2}=\frac{1}{64}\left(S_{1}-S_{3}\right)^{2} \geq \frac{1}{64}\left(\sqrt{3}-\frac{1}{3 \sqrt{3}}\right)^{2}=\frac{1}{27}, \end{aligned}\]
k.b.j.