$a,b,c>0$ and $abc=1$; prove $\sum_{cyc}\frac1{(b+1)^2}+\frac1{a+b+c+1}\ge1$

Question

Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$\sum_{cyc}\frac1{(b+1)^2}+\frac1{a+b+c+1}\ge1$$

Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that $$\displaystyle \sum_{cyc}\dfrac{1}{(b + 1)^2} \ge \displaystyle \sum_{cyc}\dfrac{1}{(c + 1)(a + 1)} = \sum_{cyc}\dfrac{1}{x - b + ca}$$ That means $$\displaystyle \sum_{cyc}\dfrac{1}{(b + 1)^2} + \dfrac{1}{a + b + c + 1} \ge \dfrac{16}{bc + ca + ab + 3x + 1}$$ $$\ge \dfrac{48}{(x - 1)^2 + 9x + 3} = \dfrac{48}{x^2 + 7x + 4}$$ That was illogical.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.

Thus, $$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{1}{a+b+c+1}-1=$$ $$=\sum_{cyc}\frac{1}{(a+1)^2}+\frac{2}{\prod\limits_{cyc}(a+1)}-1+\frac{1}{a+b+c+1}-\frac{2}{\prod\limits_{cyc}(a+1)}=$$ $$=\frac{a^2+b^2+c^2-3}{\prod\limits_{cyc}(a+1)^2}+\frac{\sum\limits_{cyc}(ab-a)}{(a+b+c+1)\prod\limits_{cyc}(a+1)}.$$ Thus, we need to prove that $$(a^2+b^2+c^2-3)(a+b+c+1)+\sum_{cyc}(ab-a)\prod_{cyc}(a+1)\geq0$$ or $$3v^4+9u^3w-6uv^2w-5uw^3-w^4\geq0,$$ which is true because $u\geq v\geq w$.

Answer 2

Hint: Substituting $$a=\frac{x}{y},b=\frac{y}{z}c=\frac{z}{x}$$ so we get

$${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3 \,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3\,{x}^{3}{y}^{ 4}{z}^{2}-3\,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{ 2}-3\,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y} ^{4}{z}^{4}+{y}^{3}{z}^{6} \geq 0$$ This is true by AM-GM.

Answer 3

Another solution.

Since $$\prod_{cyc}(a-1)^2=\prod_{cyc}((a-1)(b-1))\geq0,$$ we can assume that $(a-1)(b-1)\geq0,$ which gives $$a+b\leq ab+1=\frac{1}{c}+1.$$

Also, we have $$\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2}-\frac{1}{ab+1}=\frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}\geq0$$ Id est, $$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{1}{a+b+c+1}\geq\frac{1}{ab+1}+\frac{1}{(c+1)^2}+\frac{1}{ab+1+c+1}=$$ $$=\frac{1}{\frac{1}{c}+1}+\frac{1}{(c+1)^2}+\frac{1}{\frac{1}{c}+1+c+1}=\frac{c}{c+1}+\frac{1}{(c+1)^2}+\frac{c}{(c+1)^2}=1.$$