\(\dfrac{2xy}{x^2+4y^2}+\dfrac{y^2}{3x^2+2y^2}\le\dfrac{3}{5}\)
<=> \(\left(\dfrac{2}{5}-\dfrac{2xy}{x^2+4y^2}\right)+\left(\dfrac{1}{5}-\dfrac{y^2}{3x^2+2y^2}\right)\ge0\)
<=> \(\dfrac{2x^2+8y^2-10xy}{x^2+4y^2}+\dfrac{3x^2+2y^2-5y^2}{3x^2+2y^2}\ge0\)
<=> \(\dfrac{2\left(x-4y\right)\left(x-y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)\left(x-y\right)}{3x^2+2y^2}\ge0\)
<=> \(\left(x-y\right)\left[\dfrac{2\left(x-4y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)}{3x^2+2y^2}\right]\ge0\) (1)
Xét \(\dfrac{2\left(x-4y\right)}{x^2+4y^2}+\dfrac{3\left(x+y\right)}{3x^2+2y^2}=\dfrac{2\left(x-4y\right)\left(3x^2+2y^2\right)+3\left(x+y\right)\left(x^2+4y^2\right)}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\)
= \(\dfrac{9x^3+16xy^2-21x^2y-4y^3}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}=\dfrac{\left(x-y\right)\left(3x-2y\right)^2}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\)
(1) <=> \(\dfrac{\left(x-y\right)^2\left(3x-2y\right)^2}{\left(x^2+4y^2\right)\left(3x^2+2y^2\right)}\ge0\) (luôn đúng)
=> \(A\le\dfrac{3}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=\dfrac{2}{3}y\end{matrix}\right.\)
