# Prove $Σ_{cyc}\frac{1}{\left(a+1\right)^2}+\frac{1}{a+b+c+1}\ge 1$

Let $a,b,c>0$ such that $abc=1$. Prove the inequality $$\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}+\frac{1}{a+b+c+1}\ge 1$$

I proved $\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\ge \frac{3}{4}$ and i also prove $\frac{1}{a+b+c+1}\ge \frac{1}{4}$ but unsuccessed

• Not sure what you mean. If you proved the 2 inequalities, you can just add them to get the result. – Shirish Kulhari Mar 30 '18 at 11:58
• I think that he meant that he tried to prove those two inequalities, but failed. And he must have failed, because the AM-GM inequality tells us that, in fact, $$\frac{1}{a + b + c + 1} \leq \frac{1}{3\sqrt[3]{abc} + 1} = \frac{1}{4}.$$ – Dylan Mar 30 '18 at 12:00

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$.
Thus, we need to prove that $$\frac{\sum\limits_{cyc}(ab+a+b+1)^2}{\prod\limits_{cyc}(a+1)^2}+\frac{w}{3u+w}\geq1$$ or $$\frac{\sum\limits_{cyc}(a^2b^2+2a^2+1+2a^2b+2a^2c+2ab+2ab+4a)}{(2+a+b+c+ab+ac+bc)^2}\geq\frac{3u}{3u+w}$$ or $$\frac{9v^4-6uw^3+18u^2w^2-12v^2w^2+3w^4+18uv^2w-6w^4+12v^2w^2+12uw^3}{(2w^2+3uw+3v^2)^2}\geq\frac{3u}{3u+w}$$ or $$\frac{3v^4+6uv^2u+6u^2w^2+2uw^3-w^4}{(2w^2+3uw+3v^2)^2}\geq\frac{u}{3u+w}$$ or $$(3v^4+9u^3w-6uv^2w-5uw^3-w^4)w\geq0,$$ which is true because $u\geq v\geq w$ and $v^4\geq uw^3$.