ta có:
\(VT+4=\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ab+b^2\right)=\left(a-d\right)^2+\left(b-d\right)^2+\left(a-b\right)^2\)theo AM-GM:\(\left(a-d\right)^2+\left(b-c\right)^2\ge2\left(a-d\right)\left(b-c\right)=2\)
và \(\left(a-b\right)^2\ge0\)
do đó \(VT+4\ge2\Leftrightarrow VT\ge2\)
Dấu = xảy ra khi a=b=1;c=d=0 ...