\(\frac{a}{8}+\frac{a}{8}+\frac{1}{a^2}\ge3\sqrt[3]{\frac{a^2}{8.8.a^2}}\Rightarrow\frac{a}{4}+\frac{1}{a^2}\ge\frac{3}{4}\)
\(a\ge2\Rightarrow\frac{3a}{4}\ge\frac{3.2}{4}=\frac{3}{2}\)
Cộng vế với vế \(\Rightarrow P\ge\frac{3}{4}+\frac{3}{2}=\frac{9}{4}\)
Dấu "=" xảy ra <=>a=2