Mongolian Mathematical Olympiad

$$(a+\sqrt{b}{)}^2=a^2+2a\sqrt{b}+b\le a^2+a(b+1)+b=(a+b)(a+1).$$ By Cauchy's mean theorem $\sqrt{a^2-ab+b^2}\ge \sqrt{a^2-\frac{a^2+b^2}{2}+b^2}=\sqrt{\frac{a^2+b^2}{2}}\ge \frac{a+b}{2}$ from where we get $\frac{(a+\sqrt{b}{)}^2}{\sqrt{a^2-ab+b^2}}\ge \frac{(a+b)(a+1)}{\frac{a+b}{2}}=2(a+1)$. It implies $$\sum _{cyc}\frac{(a+\sqrt{b}{)}^2}{\sqrt{a^2-ab+b^2}}\ge 2(a+1)+2(b+1)+2(c+1)+2(d+1)=16$$

and we have done. Equality holds when $a=b=c=d=1$.