Reāliem skaitliem \(x\) un \(y\) ir spēkā vienādība \(\frac{x+y}{x-y}+\frac{x-y}{x+y}=7\). Pierādīt, ka \(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}>\sqrt{14}\).
Reizinot dotās vienādības abas puses \(\operatorname{ar}\left(x^{2}-y^{2}\right) \neq 0\), iegūstam:
\[\begin{gathered} (x+y)^{2}+(x-y)^{2}=7\left(x^{2}-y^{2}\right) \\ 2 x^{2}+2 y^{2}=7 x^{2}-7 y^{2} \quad \Rightarrow \quad y^{2}=\frac{5}{9} x^{2} \end{gathered}\]
levietojam iegūto sakarību pierādāmās nevienādības kreisās puses izteiksmē:\[\begin{aligned} \frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\frac{\frac{14}{9} x^{2}}{\frac{4}{9} x^{2}}+\frac{\frac{4}{9} x^{2}}{\frac{14}{9} x^{2}} & =\frac{7}{2}+\frac{2}{7}=\frac{53}{14}=\sqrt{\left(3 \frac{11}{14}\right)^{2}}=\sqrt{\left(4-\frac{3}{14}\right)^{2}}=\sqrt{16-\frac{24}{14}+\frac{9}{196}}= \\ & =\sqrt{14 \frac{2}{7}+\frac{9}{196}}>\sqrt{14}. \end{aligned}\]
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\[\frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\frac{7}{2}+\frac{2}{7}=\frac{53}{14}=\sqrt{\frac{53^{2}}{14^{2}}}=\sqrt{\frac{2809}{196}}=\sqrt{14 \frac{65}{196}}>\sqrt{14}.\]