Lời giải:
Áp dụng BĐT Cauchy ngược dấu:
\(\frac{2y+2z-x}{x}.3\leq \left(\frac{\frac{2y+2z-x}{x}+3}{2}\right)^2=\left(\frac{y+x+z}{x}\right)^2\)
\(\Rightarrow \frac{2y+2z-x}{x}\leq \frac{1}{3}\left(\frac{x+y+z}{x}\right)^2\)
\(\Rightarrow \sqrt{\frac{x}{2y+2z-x}}\geq \frac{\sqrt{3}x}{x+y+z}\)
Hoàn toàn tương tự:
\(\sqrt{\frac{y}{2x+2z-y}}\geq \frac{\sqrt{3}y}{x+y+z}; \sqrt{\frac{z}{2x+2y-z}}\geq \frac{\sqrt{3}z}{x+y+z}\)
Cộng theo vế những BĐT trên ta có:
\(\Rightarrow P\geq \frac{\sqrt{3}(x+y+z)}{x+y+z}=\sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\). Dấu "=" xảy ra khi \(x=y=z\)