đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)

ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)

\(\Leftrightarrow x^5-x^2\ge3x-3\)

cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)

\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)

\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)

áp dụng bunhia ta có:

\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)

cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)

đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)

\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)

\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)

\(\Rightarrow P\le1\)

đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)

\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)

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

\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)

\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)

\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)

\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)

\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)