Đáp án: $B\ge \sqrt{17}$
Giải thích các bước giải:
Ta có:
$B=(\dfrac1x+\dfrac1y)\cdot \sqrt{1+(xy)^2}$
$\to B\ge 2\cdot\sqrt{\dfrac1x\cdot\dfrac1y}\cdot \sqrt{1+(xy)^2}$
$\to B\ge 2\cdot\sqrt{\dfrac1{xy}}\cdot \sqrt{1+(xy)^2}$
$\to B\ge 2\sqrt{\dfrac1{xy}+xy}$
Đặt $xy=t$
$\to B\ge 2\sqrt{\dfrac1t+t}$
Ta có:
$1\ge x+y\ge 2\sqrt{xy}$
$\to \sqrt{xy}\le \dfrac12$
$\to xy\le\dfrac14$
$\to t\le \dfrac14$
Lại có:
$\dfrac1t+t=\dfrac{15}{16t}+(\dfrac1{16t}+t)$
$\to \dfrac1t+t\ge \dfrac{15}{16\cdot \dfrac14}+2\sqrt{\dfrac1{16t}\cdot t}$
$\to \dfrac1t+t\ge \dfrac{17}{4}$
$\to B\ge 2\cdot \sqrt{\dfrac{17}{4}}$
$\to B\ge \sqrt{17}$
Dấu = xảy ra khi $x=y=\dfrac12$
