# 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 $$ab+bc+ca=(b+c)a+bc: contradiction. Now we have $$(\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.

• Please edit your question to show the proof that you found. With respect to the (somewhat arbitrary) MathSE standards discussed here, this will allow mathSE reviewers to post alternative solutions. Feb 9, 2022 at 22:52
• @user2661923 sorry just realized the proof I found actually does not work for case b, as updated in the question. Feb 9, 2022 at 23:58
• +1 to your question, after the edit. Feb 10, 2022 at 0:30
• There are solutions here. Feb 10, 2022 at 23:27
• @CalvinLin which reply on the thread is the proof? I really would like to digest but the thread is quite messy Feb 12, 2022 at 7:37

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:

1. $$c = 1$$:

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

1. $$0< c < 1$$:

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.

• amazing. thank you very much. may I ask, how did you figure out this? there are several steps not obvious to me at all. is it that i just have to work on all such inequalities such as artofproblemsolving.com/community/c6h1183059p5735328 to get familiar with the results? or there are actually some tricks? Feb 13, 2022 at 17:16
• @athos I think the way here (finding the bounds of $ab$) to deal with the condition $a + b + c = ab + bc + ca$ works for some of such inequalities. In detail, for a symmetric $f(a,b,c)$, first find a bound $f(a,b,c)\ge g(ab, c)$ for some $g$ (with some assumption such as $c = \min(a,b,c)$), then use the bound $ab \ge \cdots$. Feb 14, 2022 at 1:02
• @athos By the way, letting $a = x^2, b = y^2, c = z^2$, we can use pqr method ($p = x + y + z, q = xy + yz + zx, r = xyz$). But perhaps it is quite complicated. Feb 14, 2022 at 1:04
• @athos The main part of the proof is (1) and (2). I gave one way to prove (1) based (2). Alternatively, one may directly insert (2) into (1) at the beginning to get an inequality of one variable $F(c) \ge 0$ for all $0 < c \le 1$. Then prove it. Feb 14, 2022 at 1:08
• thank you for sharing the insight! Feb 15, 2022 at 8:01