Ấp dụng BĐT $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq \frac{(a+b+c)^2}{x+y+z}$
Dấu $"="\Leftrightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}$
Xét $\frac{9}{4a^2+2}+\frac{9}{4b^2+2}+\frac{16}{8ab}\geq \frac{(3+3+4)^2}{4(a^2+2ab+b^2)+4}=\frac{10^2}{20}=5$$\Rightarrow \frac{1}{4a^2+2}+\frac{1}{4b^2+2}+\frac{2}{9ab}\geq \frac{5}{9}$
$\frac{1}{ab}\geq \frac{1}{\frac{(a+b)^2}{4}}=1\Rightarrow \frac{18133}{3ab}\geq \frac{18133}{3}$
Cộng 2 vế ta dc $P\geq \frac{6046}{3}$