Ta có: \(x+y+z=0\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{matrix}\right.\)
Đặt \(A=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}+3\)
Thay \(x=-\left(y+z\right),\) \(y=-\left(z+x\right),\) \(z=-\left(x+y\right)\) vào A, ta có:
\(A=\frac{y+z}{-\left(y+z\right)}+\frac{z+x}{-\left(z+x\right)}+\frac{x+y}{-\left(x+y\right)}+3\)
\(\Leftrightarrow A=\left(-1\right)+\left(-1\right)+\left(-1\right)+3\)
\(\Leftrightarrow A=-3+3\)
\(\Leftrightarrow A=0\) ( ĐPCM )
ta có:
\(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3\)
=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
mà x+y+z=0
\(\Rightarrow\)dpcm
