ÁD BĐT AM-GM cho 2 số dương ta đc:$x^{2}(y+z)\geq2x^{2}\sqrt{yz}=2x^{2}\sqrt{\frac{1}{x}}=2x\sqrt{x}$
$\Rightarrow P\geq\Sigma \frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}$
Đặt $a=x\sqrt{x};b=y\sqrt{y};c=z\sqrt{z}\Rightarrow abc=1$
Ta có:$P\geq \Sigma \frac{2a}{b+2c}=2\left[ {\Sigma \frac{a^{2}}{a(b+2c)}} \right]\geq \frac{2(a+b+c)^{2}}{a(b+2c)+b(c+2a)+c(a+2b)}$
=$\frac{2(a+b+c)^{2}}{3(ab+bc+ca)}\geq2$
Dấu''='' xra$\Leftrightarrow x=y=z=1$