Ta có:
$(b+c)(\dfrac{1}{b}+\dfrac{1}{c})\le\dfrac{(a+d)^2}{ad}$
$\Leftrightarrow \dfrac{b}{c}+\dfrac{c}{b}+2\le\dfrac{a}{d}+\dfrac{d}{a}+2$
$\Leftrightarrow \dfrac{b}{c}-\dfrac{d}{a}+\dfrac{c}{b}-\dfrac{a}{d}\le0$
$\Leftrightarrow \dfrac{ab-cd}{ca}+\dfrac{cd-ba}{bd}\le0$
$\Leftrightarrow \dfrac{(ab-cd)(bd-ca)}{abcd}\le0$, đúng do $0<a<b<c<d$.