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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.

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@Stefan4024: Where did you get the inequality from? Can you please tell us the source? – Aryabhata Apr 3 '13 at 5:06
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… for an example. – user27126 Apr 3 '13 at 5:17
@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 3 '13 at 13:58
@Aryabhata I get this inequality from a local math magazine. Any reason why it's important? – Stefan4024 Apr 3 '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 3 '13 at 16:51
up vote 1 down vote accepted

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 $(a,b,c)\in\{(0,1,1),(1,0,1),(1,1,0)\}$.

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Good one, mate. An interesting idea – Stefan4024 Jun 29 '13 at 23:16

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