Bất đẳng thức $\Leftrightarrow \frac{2}{5} - \frac{2ab}{a^2+4b^2}+\frac{1}{5} - \frac{b^2}{3a^2+2b^2}\geqslant 0 $⇔2a2−10ab+8b2a2+4b2+3a2−3b23a2+2b2⩾
\Leftrightarrow \frac{2(a-b)(a-4b)}{a^2+4b^2}+ \frac{3(a-b)(a+b)}{3a^2+2b^2}\geqslant 0
\Leftrightarrow (a-b)[2(a-4b)(3a^2+2b^2) + 3(a+b)(a^2+4b^2)]\geqslant 0
\Leftrightarrow (a-b)(9a^3 -21a^2b+16ab^2-4b^3)\geqslant 0 \Leftrightarrow (a-b)^2(3a-2b)^2\geqslant 0
=> ĐPCM . Dấu "=" xảy ra khi a=b hoặc a=\frac{2}{3}b