câu 9
gt $\Leftrightarrow 5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz =0$
$yz\leq \frac{1}{4} (y+z)^{2} ;y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$
$\Rightarrow 18yz-5(y^{2}+z^{2})\leq 2(y+z)^{2} \Rightarrow 5x^{2}-9x(y+z)\leq 2(y+z)^{2}$
$\Leftrightarrow (x-2(y+z))(5x+y+z)\leq0$
$\Rightarrow x\leq 2(y+z)$
P=$\frac{x}{y^{2}+z^{2}} -\frac{1}{(x+y+z)^{3}} \leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}}$
$\leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$ Đặt $\frac{1}{y+z}=t$
$\Rightarrow P\leq 4t-\frac{1}{27}t^{3}$
Ta sẽ chứng minh $4t-\frac{1}{27}t^{3} \leq 16$
$\Leftrightarrow \frac{(t-6)^2(t+12)}{27} \ge0$ (đúng $\forall t >0$)
$\Rightarrow P \le 16$
dấu '=" $\Leftrightarrow x=\frac{1}{3};y=z=\frac{1}{12}$