\(P=\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
\(=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)
\(=\frac{2\sqrt{1.\left(x-1\right)}}{2x}+\frac{2\sqrt{2.\left(y-2\right)}}{2y\sqrt{2}}+\frac{2\sqrt{3.\left(z-3\right)}}{2z\sqrt{3}}\)
\(\le\frac{1+x-1}{2x}+\frac{2+y-2}{2y\sqrt{2}}+\frac{3+z-3}{2z\sqrt{3}}\)(cái này của BĐT cô-si thì phải)
\(=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}1=x-1\\2=y-2\\3=z-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
Vậy \(Min_{bt}=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}\) khi \(\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
