$latex \sqrt{a^2+b^2}+\sqrt{a^2+b^2-2a+1}+\sqrt{a^2+b^2-2b+1}+\sqrt{a^2+b^2-2a-2b+2}\geq 2\sqrt2$

Solusi
Sederhanakan ruas kiri menjadi:

$latex \sqrt{a^2+b^2}+\sqrt{(1-a)^2+b^2}+\sqrt{a^2+(1-b)^2}+\sqrt{(1-a)^2+(1-b)^2}$

Perhatikan bahwa $latex a,b,1-a,1-b\in\mathbb{R}^{+}$
Maka, kita dapat menggunakan QM-AM inequality. Didapat:

$latex \sqrt{a^2+b^2}+\sqrt{(1-a)^2+b^2}+\sqrt{a^2+(1-b)^2}+\sqrt{(1-a)^2+(1-b)^2}\geq\frac{a+b}{\sqrt2}+\frac{1-a+b}{\sqrt2}+\frac{a+1-b}{\sqrt2}+\frac{1-a+1-b}{\sqrt2}=\frac{4}{\sqrt2}=2\sqrt2$ [Q.E.D.]