Reāliem skaitliem \(x\) un \(y\) ir spēkā vienādība \(\frac{x+y}{x-y}+\frac{x-y}{x+y}=7\). Pierādīt, ka \(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}>\sqrt{14}\).
Vienādojot saucējus izteiksmē \(\frac{x+y}{x-y}+\frac{x-y}{x+y}\), iegūstam, ka
\[\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{(x+y)^{2}+(x-y)^{2}}{(x+y)(x-y)}=\frac{x^{2}+2 x y+y^{2}+x^{2}-2 x y+y^{2}}{x^{2}-y^{2}}=2 \cdot \frac{x^{2}+y^{2}}{x^{2}-y^{2}}\]
Tātad, ja \(\frac{x+y}{x-y}+\frac{x-y}{x+y}=7\), tad \(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{7}{2}\) un\[\frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\frac{7}{2}+\frac{2}{7}=\frac{53}{14}=\sqrt{\frac{53^{2}}{14^{2}}}=\sqrt{\frac{2809}{196}}=\sqrt{14 \frac{65}{196}}>\sqrt{14}.\]