Ấp dụng BĐT a2x+b2y+c2z≥(a+b+c)2x+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}