Đáp án: $P\ge\sqrt{17}$
Giải thích các bước giải:
Ta có:
$P=(\dfrac1x+\dfrac1y)\sqrt{1+x^2y^2}$
$\to P\ge 2\sqrt{\dfrac1x\cdot\dfrac1y}\cdot \sqrt{1+x^2y^2}$
$\to P\ge 2\sqrt{\dfrac1{xy}}\cdot \sqrt{1+x^2y^2}$
$\to P\ge \dfrac2{xy}\cdot \sqrt{1+x^2y^2}$
$\to P\ge 2\sqrt{\dfrac1{xy}+xy}$
$\to P\ge 2\sqrt{\dfrac{15}{16xy}+\dfrac1{16xy}+xy}$
$\to P\ge 2\sqrt{\dfrac{15}{4(x+y)^2}+2\sqrt{\dfrac1{16xy}\cdot xy}}$
$\to P\ge 2\sqrt{\dfrac{15}{4\cdot 1^2}+\dfrac12}$
$\to P\ge\sqrt{17}$
Dấu = xảy ra khi $x=y=\dfrac12$
