Let $k>0~$ be fixed. Find $$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$ over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$ ================== In case we have 2 positive numbers $a+b=ab$ results that $a, b > 1$ and $a= \dfrac{b}{b-1}$, so the minimum of the expression $ab$ is 4 b...