Given $a,b,c>0$, $a+b+c=ab+bc+ca$, prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}\ge 2$

Question

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.

Answer 1

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

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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:

1. $c = 1$:

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

2. $0< c < 1$:

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.