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If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds:

$$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$

I've tried two aproaches, but it seems like both doesn't work. Here they are:

Cauchy-Scwarz inequality

I've tried using the following formula of Cauchy-Scwarz:

$$\frac{x_1^2}{a_1} + \frac{x_2^2}{a_2} + \frac{x_3^2}{a_3} \ge \frac{(x_1 + x_2 + x_3)^2}{a_1 + a_2 + a_3}$$

And i get:

$$ LHS \ge \frac{(1+1+1)^2}{2a + 2bc + 2b + 2ca + 2c + 2ba + 3}$$

$$ LHS \ge \frac{3^2}{2(a+b+c) + 5}$$

$$ LHS \ge \frac{9}{2(a+b+c) + 5}$$

Now to prove that the RHS is bigger than or equal to 1.

$$ \frac{9}{2(a+b+c) + 5} \ge 1$$

$$ 9 \ge 2(a+b+c) + 5$$

$$ 2 \ge a+b+c$$

And her I'm stuck, what can I do now?

AM - GM inequality

This try doesn't even stand a chance, but I'll still post something, because somebody can recieve an idea.

$$2a + 2bc \ge 2\sqrt{4abc}$$ $$2a + 2bc + 1 \ge 4\sqrt{abc} + 1$$ $$\frac{1}{2a + 2bc + 1} \le \frac{1}{4\sqrt{abc} + 1}$$

And for oher 2 fraction i get the same and if I add them I get:

$$\frac{3}{4\sqrt{abc} + 1} \ge LHS \ge 1$$

Now even if i prove that: $\frac{3}{4\sqrt{abc} + 1} \ge 1$ si true, that doesn't mean the original inequality holds.

Michael Rozenberg
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Stefan4024
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    @Stefan4024: Where did you get the inequality from? Can you please tell us the source? – Aryabhata Apr 03 '13 at 05:06
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    Your attempts fail because they don't respect when equality happens: which is $a = 0$ and $b=c=1$ (and its symmetric variants), rather than $a=b=c$. You may try to use $uvw$ method, which is well suited for this kind of inequality. See http://math.stackexchange.com/questions/326557/equal-variables-problem-in-three-variables/329172#329172 for an example. –  Apr 03 '13 at 05:17
  • @Aryabhata Agreed. $(a, b, c) = (100, \frac {1}{100}, 0)$ would be a cleaner example, which shows that it is unbounded. I forgot to check the boundary condition properly, because it became 0 at $(0, 0, 0)$. – Calvin Lin Apr 03 '13 at 13:58
  • @Aryabhata I get this inequality from a local math magazine. Any reason why it's important? – Stefan4024 Apr 03 '13 at 16:01
  • @Stefan4024: Only reason it is important, is to know it is solvable (and not something someone just guessed is true), and to get an idea of the "difficulty" expected. In mathematics, it can be easy to think up of hard problems (like $x^n + y^n = z^n$ :-)). Knowing where the problem came from will help answerers narrow down approaches to the solution (or knowing that it might be futile to try and solve it). – Aryabhata Apr 03 '13 at 16:51
  • @CalvinLin: Yes, now that I see $0$ is allowed :-) – Aryabhata Apr 03 '13 at 17:00
  • @Sanchez the link you provide in your other answer about the $uvw$ method has as a solved example for demonstrating the technique the inequality OP asks! – clark Apr 05 '13 at 13:12
  • @clark, yes. It is also applicable in this question since it is clear that if we keep $a+b+c$ and $ab+bc+ca$ fixed, we will need to find the maximum of quadratic expression in terms of $abc$. By $uvw$ method, we may assume that two of the variables are equal or one of them is 0. –  Apr 05 '13 at 14:17

3 Answers3

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If none of the terms $a+bc$, $b+ca$, and $c+ab$ is greater than one, then each summand of the LHS is at least $\frac{1}{3}$, and thus the inequality holds. It remains to show the inequality also holds when at least one of $a+bc$, $b+ca$, or $c+ab$ is strictly greater than one. WOLOG, suppose that $c+ab>1$. Since $c=\frac{1-ab}{a+b}$, we must have $a+b<1$. Furthermore, from Cauchy-Schwarz inequality we have $$\frac{1}{2a+2bc+1}+\frac{1}{2b+2ca+1}\geq \frac{4}{2a+2bc+1+2b+2ca+1}\\=\frac{2}{2+a+b-ab}.$$ Therefore, it suffices to show that$$\frac{2}{2+a+b-ab}+\frac{1}{2c+2ab+1} = \frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\ \geq 1, $$ holds for all $a+b<1$. We have $$A:=\frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\=\frac{4+4(a+b)-4ab+3ab(a+b)+(a+b)^2}{(2+a+b-ab)(2+a+b-2ab+2ab(a+b))}.$$ For compactness we can let $s=a+b$ and $p=ab$ and obtain $$A=\frac{4+4s-4p+3sp+s^2}{(2+s-p)(2+s-2p+2sp)}\\=\frac{4+4s-4p+3sp+s^2}{4+4s-6p+3sp+s^2+2s^2p-2p^2+2sp^2}.$$ Hence, we need to show that $$2p\geq 2s^2p-2p^2+2sp^2,$$ which is equivalent to $$2p(1-s)(1+p+s)\geq 0.$$ This inequality is valid because by assumption $s<1$. Equation cannot hold in the last case, since we need to have $p=0$ which implies $a$ or $b$ is zero, but the Cauchy-Schwarz we applied above cannot hold for these values considering $a+b<1$. The only remaining possibility for having equation is in the first case where we should have $a+bc=b+ca=a+bc=1$ which reduces to $(a,b,c)\in\{(0,1,1),(1,0,1),(1,1,0)\}$.

S.B.
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1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality it's $f(w^3)\geq0$, where $f(w^3)=-8w^6+A(u,v^2)w^3+B(u,v^2)$.

But $f$ is a concave function, which says that it's enough to prove our inequality for an extremal value of $w^3$ wich happens in the following cases.

  1. $w^3=0$.

Let $c=0$. Hence, $ab=1$ and we need to prove that $$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{3}\geq1$$ or $$\frac{a+b+1}{2(a+b)+5}\geq\frac{1}{3}$$ or $$a+b\geq2,$$ which is AM-GM: $a+b\geq2\sqrt{ab}=2$;

  1. Two variables are equal.

Let $b=a$. Hence, $a\neq0$ and $c=\frac{1-a^2}{2a}$, where $0<a<1$

and after these substitutions we need to prove that $$a^2(1-a)(1+3a-2a^2)\geq0.$$ Done!

Michael Rozenberg
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0

Supplement to @S.B.'s nice solution.

Using $a + b < 1$, we have $$\frac{2}{2 + a + b - ab} + \frac{a + b}{2 + a + b - 2ab(1 - a - b)} \ge \frac{2}{2 + a + b} + \frac{a + b}{2 + a + b} = 1.$$

We are done.

River Li
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