Đáp án: $P\ge \dfrac{63}{2}$
Giải thích các bước giải:
Ta có:
$P=\dfrac3{4x^2+9y^2}+\dfrac{5}{xy}$
$\to P=\dfrac3{(2x)^2+(3y)^2}+\dfrac{30}{2x\cdot 3y}$
Đặt $2x=a, 3y=b$
$\to a+b\le 2$
Khi đó $P=\dfrac{3}{a^2+b^2}+\dfrac{30}{ab}$
$\to P=(\dfrac{3}{a^2+b^2}+\dfrac3{2ab})+\dfrac{57}{2ab}$
$\to P=3(\dfrac{1}{a^2+b^2}+\dfrac1{2ab})+\dfrac{114}{4ab}$
$\to P\ge 3\cdot \dfrac{4}{a^2+b^2+2ab}+\dfrac{114}{(a+b)^2}$
$\to P\ge 3\cdot \dfrac{4}{(a+b)^2}+\dfrac{114}{(a+b)^2}$
$\to P\ge \dfrac{126}{(a+b)^2}$
$\to P\ge \dfrac{126}{2^2}$
$\to P\ge \dfrac{63}{2}$
Dấu = xảy ra khi $a=b=1\to 2x=3y=1\to x=\dfrac12,y=\dfrac13$
