Fie x_1, x_2, ..., x_n \in \R_+, \space \forall n \in \N, \space n \geq 2 . Să se arate că:
\Big(1+\frac{x_2}{x_1}\Big)^2 + \Big(1+\frac{x_3}{x_2}\Big)^2 +...+\Big(1+\frac{x_n}{x_{n-1} }\Big)^2 +\Big(1+\frac{x_1}{x_{n}}\Big)^2 \geq 4 \cdot nPentru demonstrarea inegalității vom considera următoarea inegalitate Minkovski:
\\(\forall)a_i, b_i\in\R_+,\; i= \overline {1,n} \\[1em]
\sqrt[n]{a_1\cdot a_2 \cdot ... \cdot a_n} + \sqrt[n]{b_1\cdot b_2 \cdot ... \cdot b_n}\leq \\\;\\ \leq \sqrt[n]{(a_1+b_1)\cdot ... \cdot(a_n+b_n)}Știm că: x_1, x_2, ..., x_n \in \R_{+} deci \frac{x_i}{x_j} \in \R_{+}; \forall i, j= \overline {1,n} atunci:
\sqrt[n]{1\cdot 1\cdot ... \cdot 1\cdot 1} + \sqrt[n]{\frac{\bcancel{x_2}}{x_1}\cdot\frac{\bcancel{x_3}}{\bcancel{x_2}} \cdot ... \cdot \frac{\bcancel{x_n}}{\bcancel{x_{n-1}}} \cdot\frac{x_1}{\bcancel{x_{n}}}}\leq \\\;\\ \leq \sqrt[n]{(1+\frac{x_2}{x_1})\cdot(1+\frac{x_3}{x_2}) \cdot ... \cdot (1+\frac{x_n}{x_{n-1}})\cdot(1+\frac{x_1}{x_{n}})} \iff \\[2em]
\sqrt[n]{1}+\sqrt[n]{1} \leq \sqrt[n]{(1+\frac{x_2}{x_1})\cdot(1+\frac{x_3}{x_2}) \cdot ... \cdot (1+\frac{x_n}{x_{n-1}})\cdot(1+\frac{x_1}{x_{n}})} \iff \\[2em]
2 \leq \sqrt[n]{(1+\frac{x_2}{x_1})\cdot(1+\frac{x_3}{x_2}) \cdot ... \cdot (1+\frac{x_n}{x_{n-1}})\cdot(1+\frac{x_1}{x_{n}})}
Dar știm de-asemena din inegalitatea mediilor, că media geometrică este mai mică decât media pătratică deci:
{\sqrt[n]{(1+\frac{x_2}{x_1})\cdot(1+\frac{x_3}{x_2}) \cdot ... \cdot (1+\frac{x_n}{x_{n-1}})\cdot(1+\frac{x_1}{x_{n}})}} \leq \\[2em]\\[2em]
\leq \sqrt[]{\frac{\Big(1+\frac{x_2}{x_1}\Big)^2 + \Big(1+\frac{x_3}{x_2}\Big)^2 +...+\Big(1+\frac{x_n}{x_{n-1} }\Big)^2 +\Big(1+\frac{x_1}{x_{n}}\Big)^2}{n}} Deci:
2 \leq \sqrt[]{\frac{\Big(1+\frac{x_2}{x_1}\Big)^2 + \Big(1+\frac{x_3}{x_2}\Big)^2 +...+\Big(1+\frac{x_n}{x_{n-1} }\Big)^2 +\Big(1+\frac{x_1}{x_{n}}\Big)^2}{n}} \iff \\[2em]
4\cdot n \leq \Big(1+\frac{x_2}{x_1}\Big)^2 + \Big(1+\frac{x_3}{x_2}\Big)^2 +...+\Big(1+\frac{x_n}{x_{n-1} }\Big)^2 +\Big(1+\frac{x_1}{x_{n}}\Big)^2
q.e.d
