$gt\Rightarrow 3\leq 4ab-2(a+b)\leq (a+b)^{2}-2(a+b)\Rightarrow (a+b)\geq 3$$a^{4}+b^{4}\geq \frac{1}{2}(a^{2}+b^{2})^{2}\geq \frac{1}{8}(a+b)^{4}$
$\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ do đó
$P\geq \sqrt{\frac{1}{8}(a+b)^{4}}(\frac{4}{a+b}-\frac{2}{a+b})=\frac{1}{2\sqrt{2}}.(a+b)^{2}.\frac{2}{a+b}=\frac{1}{\sqrt{2}}.(a+b)\geq \frac{3}{\sqrt{2}}$
dấu bằng khi $a=b=\frac{3}{2}$