gt$\Leftrightarrow$$5x^{2}+5(y^{2}+z^{2})-9x(y+z)-18yz=0$(1)Ta có:$yz\leq \frac{1}{4}(y+z)^{2}$;$y^{2}+z^{2}\geq \frac{1}{2}(y+z)^{2}$
$\Rightarrow$$18yz-5(y^{2}+z^{2})\leq2(y+z)^{2}$(2)
Từ(1)&(2)$\Rightarrow$$5x^{2}-9x(y+z)\leq2(y+z)^{2}$
$\Leftrightarrow$$(x-2(y+z))(5x+y+z)\leq0$
$\Rightarrow$$x\leq2(y+z)$
P$\leq \frac{2x}{(y+z)^{2}}-\frac{1}{(x+y+z)^{3}} \leq \frac{4}{y+z}-\frac{1}{27(y+z)^{3}}$
Đặt t=$\frac{1}{y+z}$$\Rightarrow$P$\leq$4t-$\frac{t^{3}}{27}$
Ta phải cm P$\leq$16
$\Leftrightarrow$$\frac{(t+12)(t-6)^{2}}{27}\geq$0(luôn đúng)
Dấu''='' xra$\Leftrightarrow$$x=\frac{1}{3};y=z=\frac{1}{12}$