Giải thích các bước giải:
$P=(\dfrac 1x+\dfrac 1y)\sqrt{1+x^2y^2}$
$\to P\ge 2\sqrt{\dfrac 1x.\dfrac 1y}\sqrt{1+x^2y^2}$
$\to P\ge 2\sqrt{\dfrac{1}{xy}+xy}$
$\to P\ge 2\sqrt{\dfrac{1}{4xy}+xy+\dfrac{3}{4xy}}$
$\to P\ge 2\sqrt{2\sqrt{\dfrac{1}{4xy}.xy}+\dfrac{3}{(x+y)^2}}$
$\to P\ge 4$