数学問題13番
計算
左辺=
\(\displaystyle\frac{x-y}{y+z}+\displaystyle\frac{y-z}{z+x}+\displaystyle\frac{z-x}{x+y} \)
\(=\displaystyle\frac{(x+z)-(y+z)}{y+z}+\displaystyle\frac{(y+x)-(z+x)}{z+x}+\displaystyle\frac{(z+y)-(x+y)}{x+y}\)
\(=\displaystyle\frac{x+z}{y+z}+\displaystyle\frac{y+x}{z+x}+\displaystyle\frac{z+y}{x+y}-3\)
\( \geq 3\sqrt[3]{\displaystyle\frac{x+z}{y+z}\cdot\displaystyle\frac{y+x}{z+x}\cdot\displaystyle\frac{z+y}{x+y}}-3=0\) ……(相加相乗)
よって問題の不等式は成立。