Problem - 4783
Let positive numbers $a$, $b$ and $c$ satisfy $a+b+c=8$. Find the minimal value of $\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9}$.
Answer
$10$
By # 4782, we have $$\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9}\ge\sqrt{(a+b+c)^2+(1+2+3)^2}=\sqrt{8^2 + 6^2}=\boxed{10}$$