For $a,b,c \ge 0$ real numbers, prove $\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \geq \frac{9}{3+ab+bc+ca}$

Question

Let be $a,b,c \ge 0$ real numbers. Prove that:

$$\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \geq \frac{9}{3+ab+bc+ca}$$

It is question 2 from here : https://artofproblemsolving.com/community/c6h1934647p13303456

I used Titu's lemma

$$\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \ge \frac{18}{3+2(a+b+c)+a^2+b^2+c^2}$$

But

$$\frac{18}{3+2(a+b+c)+a^2+b^2+c^2} \ge \frac{9}{3+ab+bc+ca}$$

$$\Leftrightarrow 3+2(ab+bc+ca) \ge 2(a+b+c)+a^2+b^2+c^2$$

this is false when $a=0, b=c=1$.

Answer 1

First we prove:

$$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2} \geq \frac{1}{1+ab}$$

From Cauchy-Schwarz:

$$(1+ab)\left(1+\frac{a}{b}\right)\geq (1+a)^2\Rightarrow \frac{1+ab}{(1+a)^2}\geq \frac{b}{a+b}$$

Similarly:

$$\frac{1+ab}{(1+b)^2}\geq \frac{a}{a+b}$$

Summing up:

$$\frac{1+ab}{(1+a)^2}+\frac{1+ab}{(1+a)^2}\geq \frac{a}{a+b}+\frac{b}{a+b}=1$$

and thus:

$$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2} \geq \frac{1}{1+ab}$$

Summing with the other two symmetric inequalities:

$$\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \geq \frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}$$

and:

$$\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca} \geq \frac{9}{3+ab+bc+ca}$$

follows from Cauchy-Schwarz (or Titu's lemma as you call it).

Answer 2

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

Now, by C-S $$\sum_{cyc}\frac{2}{(1+a)^2}=\sum_{cyc}\frac{2(b+c)^2}{(b+c)^2(1+a)^2}\geq\frac{8(a+b+c)^2}{\sum\limits_{cyc}(b+c)^2(1+a)^2}=$$ $$=\frac{72u^2}{\sum\limits_{cyc}(2a^2+2ab+2a^2b+2a^2c+4abc+2a^2b^2+2a^2bc)}=$$ $$=\frac{36u^2}{9u^2-3v^2+9uv^2+3w^3+9v^4-3uw^3}=$$ $$=\frac{12u^2}{3u^2-v^2+3uv^2+w^3+3v^4-uw^3}.$$ Thus, it's enough to prove that $$\frac{4u^2}{3u^2-v^2+3uv^2+w^3+3v^4-uw^3}\geq\frac{1}{1+v^2}$$ or $$u^2+v^2+4u^2v^2-3v^4+uw^3\geq3uv^2+w^3.$$ But by AM-GM $$u^2+v^2+4u^2v^2-3v^4+uw^3\geq2\sqrt{(u^2+v^2)(4u^2v^2-3v^4+uw^3)}.$$ Thus, it's enough to prove that $$4(u^2+v^2)(4u^2v^2-3v^4+uw^3)\geq(3uv^2+w^3)^2$$ or $$16u^4v^2-5u^2v^4-12v^4+4u^3w^3-2uv^2w^3-w^6\geq0,$$ for which it's enough to prove that: $$16u^4v^2-5u^2v^4-12v^4+uv^2w^3\geq0$$ or $$16u^4-5u^2v^2-12v^2+uw^3\geq0.$$ But by Schur $$w^3\geq4uv^2-3u^3.$$ Id est, it's enough to prove that $$16u^4-5u^2v^2-12v^2+u(4uv^2-3u^3)\geq0$$ or $$(u^2-v^2)(13u^2+12v^2)\geq0,$$ which is obvious.