Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$
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.
Solution 1:
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 $(a,b,c)\in\{(0,1,1),(1,0,1),(1,1,0)\}$.
Solution 2:
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.
- $w^3=0$.
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: $a+b\geq2\sqrt{ab}=2$;
- 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!