syk0526 wrote: Let $ x, y, z, w $ be nonzero real numbers such that $ x+y \ne 0$, $ z+w \ne 0 $, and $ xy+zw \ge 0 $. Prove that \[ \left( \frac{x+y}{z+w} + \frac{z+w}{x+y} \right) ^{-1} + \frac{1}{2} \ge \left( \frac{x}{z} + \frac{z}{x} \right) ^{-1} + \left( \frac{y}{w} + \frac{w}{y} \right) ^{-1}\] $\left( \frac{x+y}{z+w} + \frac{z+w}{x+y} \right) ^{-1} + \frac{1}{2} \ge \left( \frac{x}{z} + \frac{z}{x} \right) ^{-1} + \left( \frac{y}{w} + \frac{w}{y} \right) ^{-1}\Leftrightarrow$ $\Leftrightarrow\frac{1}{2}-\frac{xz}{x^2+z^2}+\frac{1}{2}-\frac{yw}{y^2+w^2}\geq\frac{1}{2}-\frac{(x+y)(z+w)}{(x+y)^2+(z+w)^2}\Leftrightarrow$ $\Leftrightarrow\frac{(x-z)^2}{x^2+z^2}+\frac{(y-w)^2}{y^2+w^2}\geq\frac{(x+y-z-w)^2}{(x+y)^2+(z+w)^2}$, which is true because $\frac{(x-z)^2}{x^2+z^2}+\frac{(y-w)^2}{y^2+w^2}\geq\frac{(x+y-z-w)^2}{x^2+y^2+z^2+w^2}\geq\frac{(x+y-z-w)^2}{(x+y)^2+(z+w)^2}$.

arqady wrote: $\Leftrightarrow\frac{(x-z)^2}{x^2+z^2}+\frac{(y-z)^2}{y^2+z^2}\geq\frac{(x+y-z-w)^2}{(x+y)^2+(z+w)^2}$, which is true because $\frac{(x-z)^2}{x^2+z^2}+\frac{(y-z)^2}{y^2+z^2}\geq\frac{(x+y-z-w)^2}{x^2+y^2+z^2+w^2}\geq\frac{(x+y-z-w)^2}{(x+y)^2+(z+w)^2}$. there should be $z$ replaced by $w$ in the $\frac{(y-z)^2}{y^2+z^2}$ (P.S. nice solution! )