\(H=\frac{1}{\left(x+1\right)^2+y^2+1}+\frac{1}{\left(y+1\right)^2+z^2+1}+\frac{1}{\left(z+1\right)^2+x^2+1}\)
\(\Leftrightarrow\)\(H=\frac{1}{\left(x+1\right)^2+\left(y+1\right)^2-2y}+\frac{1}{\left(y+1\right)^2+\left(z+1\right)^2-2z}+\frac{1}{\left(z+1\right)^2+\left(x+1\right)^2-2x}\)
Áp dụng BĐT AM-GM ta có:
\(H\le\frac{1}{2.\left(x+1\right)\left(y+1\right)-2y}+\frac{1}{2.\left(y+1\right)\left(z+1\right)-2z}+\frac{1}{2.\left(z+1\right)\left(x+1\right)-2x}\)
\(\Leftrightarrow H\le\frac{1}{2.\left(x+y+xy+1\right)-2y}+\frac{1}{2.\left(y+z+yz+1\right)-2z}+\frac{1}{2.\left(x+z+xz+1\right)-2x}\)
\(\Leftrightarrow H\le\frac{1}{2.\left(x+xy+1\right)}+\frac{1}{2.\left(y+yz+1\right)}+\frac{1}{2.\left(z+xz+1\right)}\)
\(\Leftrightarrow H\le\frac{1}{2}\left[\frac{xyz}{x\left(1+y+yz\right)}+\frac{1}{y+yz+1}+\frac{xyz}{xz\left(y+yz+1\right)}\right]\)
\(\Leftrightarrow H\le\frac{1}{2}\left[\frac{yz}{1+y+yz}+\frac{1}{y+yz+1}+\frac{y}{y+yz+1}\right]=\frac{1}{2}.1=\frac{1}{2}\)
Dấu " = " xảy ra <=> \(x=y=z=1\)
Vậy \(H_{max}=\frac{1}{2}\Leftrightarrow x=y=z=1\)
