Đặt \(A=\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\)
\(A=\frac{x}{\sqrt{x^2+xy+yz+zx}}+\frac{y}{\sqrt{y^2+xy+yz+zx}}+\frac{z}{\sqrt{z^2+xy+yz+zx}}\)
\(A=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+x\right)\left(y+z\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
Áp dụng bđt Cô-si ta được:\(\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{x}{2\left(x+y\right)}+\frac{x}{2\left(x+z\right)}\)
\(\frac{y}{\sqrt{\left(y+x\right)\left(y+z\right)}}\le\frac{y}{2\left(y+x\right)}+\frac{y}{2\left(y+z\right)};\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\le\frac{z}{2\left(z+x\right)}+\frac{z}{2\left(z+y\right)}\)
=>A\(\le\frac{x}{2\left(x+y\right)}+\frac{x}{2\left(x+z\right)}+\frac{y}{2\left(y+x\right)}+\frac{y}{2\left(y+z\right)}+\frac{y}{2\left(z+x\right)}+\frac{y}{2\left(z+y\right)}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)
dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
