Cách giải:
$a^2+b^2+c^2+d^2 \geq (a+b)(c+d)$
$\to 2a^2+2b^2+2c^2+2d^2 \geq 2ac+2ad+2bc+2bd$
$\to 2a^2+2b^2+2c^2+2d^2-2ac-2ad-2bc-2bd \geq 0$
$\to a^2-2ac+c^2+a^2-2ad+d^2+b^2-2bc+c^2+b^2-2bd+d^2 \geq 0$
$\to (a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2 \geq 0$ luôn đúng
Dấu "=" xảy ra khi
$\begin{cases}a=c\\a=d\\b=c\\b=d\\\end{cases}$
$\to a=b=c=d$
