ta có 

\(4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\left(x-y\right)^2\ge0\) vì thế \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)

hoàn toàn tương tự ta sẽ có 

\(P\ge\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{\sqrt{3}}{2}\left(y+z\right)+\frac{\sqrt{3}}{2}\left(x+z\right)\)

hay

\(P\ge\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

dấu bằng xảy ra khi x=y=z=1

 Ta có: \(P=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)\(=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2+\frac{1}{4}\left(y-z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2+\frac{1}{4}\left(x-y\right)^2}\)\(\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2}=\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

Đẳng thức xảy ra khi x = y = z = 1

Áp dụng giả thiết và bất đẳng thức AM - GM, ta có: \(VT=\frac{x}{\sqrt{x^2+xy+yz+zx}}+\frac{y}{\sqrt{y^2+xy+yz+zx}}+\frac{z}{\sqrt{z^2+xy+yz+zx}}\)\(=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+x\right)\left(y+z\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)\(=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}.\frac{z}{y+z}}\)\(\le\frac{\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{y+z}}{2}=\frac{3}{2}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)