Đáp án: $P\le\dfrac8{29}$
Giải thích các bước giải:
Ta có:
$P=\dfrac{2xy}{x^2+y^2+3xy}$
$\to P=\dfrac{2}{\dfrac{x}{y}+\dfrac{y}{x}+3}$
Đặt $\dfrac4y=t \to x+t\le 2$
$\to P=\dfrac{2}{\dfrac14xt+\dfrac4{xt}+3}$
Ta có:
$\dfrac14xt+\dfrac4{xt}+3=(\dfrac14xt+\dfrac1{4xt})+\dfrac{15}{4xt}+3$
$\to\dfrac14xt+\dfrac4{xt}+3\ge 2\sqrt{\dfrac14xt\cdot\dfrac1{4xt}}+\dfrac{15}{(x+t)^2}+3$
$\to\dfrac14xt+\dfrac4{xt}+3\ge \dfrac12+\dfrac{15}{2^2}+3$
$\to\dfrac14xt+\dfrac4{xt}+3\ge \dfrac{29}4$
$\to P\le \dfrac{2}{\dfrac{29}4}=\dfrac8{29}$
Dấu = xảy ra khi $x=t=1$
$\to \dfrac4y=1\to y=4$
