$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ 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...
 A: Solution for $k\ge 24$:
Let $k\ge 24$ be a real number (fixed).
Let $f(a, b, c) = \sqrt{ka + 1} + \sqrt{kb + 1} + \sqrt{kc + 1}$.
Let us prove that $f(a, b, c) \ge f(2, 2, 0) = 2\sqrt{2k + 1} + 1$
for all $a, b, c\ge 0$ with $ab + bc + ca = a + b + c > 0$.
WLOG, assume that $c = \min(a, b, c)$.
From $ab + bc + ca = a + b + c$, we have
$(1 - c)(a + b) = ab - c$. Thus, $c \le 1$.
Thus, we have $(1 - c)\cdot 2\sqrt{ab} \le ab - c$
or $(\sqrt{ab} - 1 + c)^2 \ge c^2 - c + 1$ which results
in $\sqrt{ab} \ge \sqrt{c^2 - c + 1} + 1 - c$.
We have
\begin{align*}
 f(a, b, c) &= \sqrt{\left( \sqrt{ka + 1} + \sqrt{kb + 1}\right)^2} + \sqrt{kc + 1}\\
 &= \sqrt{k(a + b) + 2 + 2\sqrt{k^2ab + k(a + b) + 1}} + \sqrt{kc + 1}\\
 &\ge \sqrt{k\cdot 2\sqrt{ab} + 2 + 2\sqrt{k^2ab + k\cdot 2\sqrt{ab} + 1}} + \sqrt{kc + 1}\\
 &= \sqrt{k\cdot 2\sqrt{ab} + 2 + 2\cdot (k\sqrt{ab} + 1)} + \sqrt{kc + 1}\\
 &= 2\sqrt{k\sqrt{ab} + 1} + \sqrt{kc + 1}\\
 &\ge 2\sqrt{k(\sqrt{c^2-c+1} + 1-c) + 1} + \sqrt{kc + 1}.
\end{align*}
It suffices to prove that, for all $c$ in $[0, 1]$,
$$2\sqrt{k(\sqrt{c^2-c+1} + 1-c) + 1} + \sqrt{kc + 1}\ge 2\sqrt{2k + 1} + 1.$$
Let $\sqrt{c^2 - c + 1} - c = u$.
Then $0 \le u \le 1$ and $c = \frac{1 - u^2}{2u + 1}$.
It suffices to prove that, for all $u$ in $[0, 1]$,
$$2\sqrt{ku + k + 1} + \sqrt{k \cdot \frac{1 - u^2}{2u + 1} + 1} \ge 2\sqrt{2k + 1} + 1$$
or (squaring both sides)
\begin{align*}
 &4\sqrt{ku + k + 1}\sqrt{k \cdot \frac{1 - u^2}{2u + 1} + 1}\\
 \ge\ & (2\sqrt{2k + 1} + 1)^2 - 4(ku + k + 1)
 - k \cdot \frac{1 - u^2}{2u + 1} - 1
\end{align*}
or (squarding both sides again)
\begin{align*}
 &16(ku + k + 1)\left(k \cdot \frac{1 - u^2}{2u + 1} + 1\right)\\
 \ge\ & \left[(2\sqrt{2k + 1} + 1)^2 - 4(ku + k + 1)
 - k \cdot \frac{1 - u^2}{2u + 1} - 1\right]^2
\end{align*}
that is
$$\frac{(1 - u)k}{(2u + 1)^2}(Au^3 + Bu^2 + Cu + D) \ge 0$$
where
\begin{align*}
 A &= 81k, \\
 B &=  -112\,\sqrt {2\,k+1}+73\,k-32, \\
 C &=  -104\,\sqrt {2\,k+1}+31\,k-16, \\
 D &= -24\,\sqrt {2\,k+1}+7\,k.
\end{align*}
It is easy to prove that $A, B, C, D \ge 0$ for all $k \ge 24$.
We are done.
A: 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:
$$\min E(a,b,c,k)=\min_{k>0}\left\{3\cdot\sqrt{k+1},1+2\cdot\sqrt{2k+1}\right\}.$$
