-Xét \(x\ge y\ge z\). Dễ cm bđt đúng
-Xét \(x\ge z\ge y\)
Đặt x=z+a, z=y+b với \(a,b\ge0\)
=>x=y+a+b
BĐT\(< =>\frac{x-y}{y\left(y+1\right)}\ge\frac{x-z}{x\left(x+1\right)}+\frac{z-x}{z\left(z+1\right)}\)
<=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)
Vì \(x\ge z\ge y=>x\left(x+1\right)\ge z\left(z+1\right)\ge y\left(y+1\right)\)
\(=>\frac{a}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)},\frac{b}{y\left(y+1\right)}\ge\frac{b}{z\left(z+1\right)}\)
=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)=>bđt cần cm đúng=>đpcm