<s>[INST] $\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ Let $k>0~$ be fixed. Find
$$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$
over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$
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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 because $\dfrac{b}{b-1} \cdot b \geqslant 4$ --- equality when $a=b=2$
Then $(\sqrt{ka+1}  + \sqrt{kb+1})^2 = kab+2+ 2\sqrt{k^2ab + kab +1} ~~~$ (I've used $a+b = ab$)
so the minimum is reached when $ab$ is minimal, that is 4 so we obtain $4k+2 + 2\sqrt{4k^2 + 4k + 1} = 4(2k+1)$
That is the minimum searched is $2\sqrt{2k+1}$.
In case of 3 numbers I wonder if an idea is to raise to power 2 and proceed in some similar way...
 [/INST] Sketch of my solution:
Set WLOG $a\leq b\leq c.$ We denote $bc=p^2$  and $b+c=2s.$ Clearly, $s\geq p\geq1.$ By AM-GM and CBS,we get:
$$\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\geq\sqrt{ka+1}+2\cdot\sqrt{\sqrt{kb+1}\cdot\sqrt{kc+1}}\geq\sqrt{ka+1}+2\cdot\sqrt{kp+1}.$$
Case 1: $p>2.$ We have:
$$E(a,b,c,k)\geq \sqrt{ka+1}+2\cdot\sqrt{kp+1}>1+2\cdot\sqrt{2k+1}.$$
Case 2: $p\in[1,2].$ According to BMO Problem 2/2019, $a\geq\frac{2-p}p,$ with equality iff $a=b=c=1$  or $a=0,b=c=2.$ Thus,
$$E(a,b,c,k)\geq \sqrt{ka+1}+2\cdot\sqrt{kp+1}\geq\sqrt{k\cdot\frac{2-p}p+1}+2\cdot\sqrt{kp+1}.$$
We now take the first derivative of the function $f(p):=\sqrt{k\cdot\frac{2-p}p+1}+2\cdot\sqrt{kp+1}$ and trivially get:
$$\min E(a,b,c,k)=\min_{k>0}\left\{3\cdot\sqrt{k+1},1+2\cdot\sqrt{2k+1}\right\}.$$
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