Given $a,b,c>0$, $a+b+c=ab+bc+ca$, prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$ Given $a,b,c>0$, $$a+b+c=ab+bc+ca$$, prove $$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$$
I tried with derivatives but haven't solved it yet.
Is there a more natural and elementary proof?
My progress with derivatives:
First $a+b+c=ab+bc+ca\le \frac{(a+b+c)^2}{3}$, so $a+b+c\ge 3$.
Then $a+b+c=ab+bc+ca \Leftrightarrow \frac1a+\frac1b+\frac1c=\frac1{ab}+\frac1{bc}+\frac1{ca}$, so similarly $\frac1a+\frac1b+\frac1c\ge 3$. Hence $ab+bc+ca\ge 3abc$.
WOLOG, we assume $a>b>c$. Obviously $a>1>c$.
We have $b+c > 1$, otherwise $ab+bc+ca=(b+c)a+bc<a+bc<a+b+c$: contradiction. Now we have $(\sqrt b +\sqrt c)^2 > b+c >1$, so $\sqrt b+ \sqrt c>1$.
With notation $s:=\sqrt a + \sqrt b + \sqrt c$ and $t:=\sqrt{abc}$, we have $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2 \Leftrightarrow (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 \ge (2+\sqrt t)^2 $, which is equivalent to $$ab+bc+ca + 2s \sqrt{t}  \ge 4 + 4\sqrt t + t$$, call it (*), now we discuss by case a) $abc \ge 1$ and case b) $abc<1$.
Case a) $abc \ge 1$. Inequation (*) holds if $$3t+6\sqrt t \sqrt[6] t\ge 4 + 4\sqrt t + t$$ holds, which  again is equivalent to $$-2t+4\sqrt t -6t^{\frac23}+4\le 0$$ , but this can be proved by computing the derivative.
Case b) $abc<1$. I haven't solved this yet.

Remark: @Calvin Lin's comment reminds me that, in 2019, I proved some similar inequalities under the condition $a + b + c = ab + bc + ca$.

WLOG, assume that $c = \min(a, b, c)$.
Using AM-GM, we have
\begin{align*}
 &\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - \sqrt{abc} - 2\\
 \ge\,& \sqrt{ab}
 + 2\sqrt{\sqrt{bc}\sqrt{ca}} - \sqrt{abc} - 2\\
 =\,& (1 - \sqrt{c})\sqrt{ab} + 2\sqrt{c}\sqrt[4]{ab} - 2. \tag{1}
\end{align*}
From $ab + bc + ca = a + b + c$, we have
$(1 - c)(a + b) = ab - c$. Thus, $c \le 1$ (easy).
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
$$\sqrt{ab} \ge \sqrt{c^2 - c + 1} + 1 - c. \tag{2}$$
Note: $\sqrt{ab} - 1 + c \le -\sqrt{c^2 - c + 1}$ is impossible, since $1 - c - \sqrt{c^2 - c + 1} < 0$ for all $c > 0$.
We split into two cases:

*

*$c = 1$:

From (2), we have $\sqrt{ab} \ge 1$.
From (1), the desired result follows.


*$0< c < 1$:

From (1), it suffices to prove that
$$(1 - \sqrt{c})\sqrt{ab} + 2\sqrt{c}\sqrt[4]{ab} - 2 \ge 0$$
which is written as
$$(1 - \sqrt c)\left(\sqrt[4]{ab} + \frac{\sqrt c}{1 - \sqrt c}\right)^2 \ge \frac{2 + c - 2\sqrt c}{1 - \sqrt c}.$$
It suffices to prove that
$$\sqrt[4]{ab} + \frac{\sqrt c}{1 - \sqrt c} \ge \frac{\sqrt{2 + c - 2\sqrt c}}{1 - \sqrt c}$$
or
$$\sqrt[4]{ab}
\ge \frac{\sqrt{2 + c - 2\sqrt c} - \sqrt c}{1 - \sqrt c}
= \frac{2}{\sqrt{2 + c - 2\sqrt c} + \sqrt c}$$
or
$$\sqrt{ab} \ge \frac{4}{(\sqrt{2 + c - 2\sqrt c} + \sqrt c)^2} = \frac{4}{2 + 2c - 2\sqrt c + 2\sqrt c \sqrt{1 + (1 - \sqrt c)^2}}.$$
It suffices to prove that
$$\sqrt{ab} \ge \frac{4}{2 + 2c - 2\sqrt c + 2\sqrt c } = \frac{2}{1 + c}.$$
Using (2), it suffices to prove that
$$\sqrt{c^2 - c + 1} + 1 - c  \ge \frac{2}{1 + c}$$
or
$$\sqrt{c^2 - c + 1} \ge \frac{1 + c^2}{1 + c}$$
or
$$c^2 - c + 1 \ge \left(\frac{1 + c^2}{1 + c}\right)^2$$
or
$$\frac{c(1 - c)^2}{(1 + c)^2}\ge 0.$$
We are done.
