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.

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. — user2661923, Feb 09 '22 at 22:52
@user2661923 sorry just realized the proof I found actually does not work for case b, as updated in the question. — athos, Feb 09 '22 at 23:58
one can do it with Lagrange multipliers in the case $abc\le 1$ by showing that under $a \ge b \ge c$ we have $(\sqrt {ab}+\sqrt {ac})^2 \ge (\sqrt a + \sqrt {abc} -\sqrt {bc})^2$ by fairly straightforward simplifications; so the inequality is true for $a >4$ and then we work in a box $1 \le a \le4, 0\le b \le 4, 0 \le c \le 1$ and find the critical points inside for the required expression under the constraint and then analyze the boundary, while the case $abc \ge 1$ can be proven from $abc \le 1$ using the substitutions $a \to 1/a$ etc and some considerations, but it is kind of ugly overall — Conrad, Feb 10 '22 at 19:02
applying Lagrange multipliers is actually quite easy, the problem is in finding a good boundary and the main issue imho is that there is a minimum for $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{abc}$ at $(1,1,1)$ but also one in the limit at $(2, 2, 0)$ and that makes the case $c$ very small (and $abc$ small) tricky — Conrad, Feb 10 '22 at 19:06
@Conrad I’m not sure if I understand your idea, do you mind expand it to an answer pls? — athos, Feb 10 '22 at 19:27
for simplicity with $a=x^2, b=y^2, c=z^2, 0 \le z \le y \le x, x \ge 1$, you need to minimize $f(x,y,z)=xy+yz+zx-xyz$ under the constraint $x^2+y^2+z^2-(xy)^2-(yz)^2-(zx)^2=0$ and that is a non singular surface in our case ($x \ge1$) so you can use Lagrange multipliers assuming you have a good boundary — Conrad, Feb 10 '22 at 19:44
@CalvinLin which reply on the thread is the proof? I really would like to digest but the thread is quite messy — athos, Feb 12 '22 at 07:37

score 1 · Accepted Answer · answered Feb 13 '22 at 01:13

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.

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 https://artofproblemsolving.com/community/c6h1183059p5735328 to get familiar with the results? or there are actually some tricks? — athos, Feb 13 '22 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 $ab \ge \cdots$. — River Li, Feb 14 '22 at 01:02
@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. — River Li, Feb 14 '22 at 01:04
@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. — River Li, Feb 14 '22 at 01:08

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

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