Prove $\frac{4}{(a+1)(b+1)(c+1)}+\frac{1}{4}\ge \frac{a}{(a+1)^2}+\frac{b}{(b+1)^2}+\frac{c}{(c+1)^2}.$

Question

Let $a,b,c>0: abc=1.$ Prove that$$\frac{4}{(a+1)(b+1)(c+1)}+\frac{1}{4}\ge \frac{a}{(a+1)^2}+\frac{b}{(b+1)^2}+\frac{c}{(c+1)^2}.$$ I've tried to use equivalent steps but it is quite complicated.

Indeed, we'll prove $$4+\frac{(a+1)(b+1)(c+1)}{4}\ge \sum_{cyc}\frac{a(b+1)(c+1)}{a+1}.$$ From here, I don't know how to continue. Can you help me?

Answer 1

Proof.

Denote$$(x,y,z)\sim \left(\frac{1}{a+1},\frac{1}{b+1},\frac{1}{c+1}\right).$$ Now, rewrite the original inequality as$$4xyz+\frac{1}{4}\ge \sum_{cyc} (x-x^2) \iff x^2+y^2+z^2+4xyz+\frac{1}{4}\ge x+y+z. \tag{*}$$ By the given condition and substitution, we get $2xyz=1-(x+y+z)+(xy+yz+zx).$

Thus, $(*)$ turns out$$x^2+y^2+z^2+2(xy+yz+xz)+\frac{9}{4}\ge 3(x+y+z),$$or$$\left(x+y+z-\frac{3}{2}\right)^2\ge 0,$$which gives completed proof. Equality holds at $a=b=c=1.$

Answer 2

Here is what I thought. First,

$$(a-1)(b-1)(c+1) = abc + ab - ac - bc + c - a - b + 1$$

then $$\sum_{cyc} (a-1)(b-1)(c+1) = 3abc + 3 - \sum_{cyc} ab - \sum_{cyc} a = 6 - \sum_{cyc} ab - \sum_{cyc} a = 8 - \prod(a+1)$$

now $$\sum_{cyc} \frac{ (a-1)(b-1)}{ (a+1)(b+1)} = \sum_{cyc} \frac{ (a-1)(b-1)(c+1)}{ (a+1)(b+1)(c+1)}=\frac{8}{\prod(a+1)} - 1$$

and

$$\sum_{cyc} \frac{(a-1)^2}{(a+1)^2} = 3-\sum_{cyc} \frac{4a}{(a+1)^2} $$

thus we only need

$$-2 \sum_{cyc} \frac{ (a-1)(b-1)}{ (a+1)(b+1)} \le \sum_{cyc} \frac{(a-1)^2}{(a+1)^2}$$

That is

$$\left(\sum_{cyc} \frac{a-1}{a+1}\right)^2 \ge 0. $$

Answer 3

SOS helps!

Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{z}{x}$ and we need to prove that: $$\frac{16xyz}{\prod\limits_{cyc}(x+y)}+1\geq4\sum_{cyc}\frac{xy}{(x+y)^2}$$ or $$3-4\sum_{cyc}\frac{xy}{(x+y)^2}\geq2-\frac{16xyz}{\prod\limits_{cyc}(x+y)}$$ or $$\sum_{cyc}\frac{(x-y)^2}{(x+y)^2}\geq\sum_{cyc}\frac{2z(x-y)^2}{\prod\limits_{cyc}(x+y)}$$ or $$\sum_{cyc}\frac{(x-y)^2((x+z)(y+z)-2z(x+y))}{(x+y)^2}\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2(x-z)(y-z)}{(x+y)^2}\geq0$$ or $$(x-y)(y-z)(z-x)\sum_{cyc}\frac{y-x}{(x+y)^2}\geq0$$ or $$\prod_{cyc}(x-y)\sum_{cyc}(y-x)(x+z)^2(y+z)^2\geq0$$ or $$\prod_{cyc}(x-y)\sum_{cyc}(y-x)(z^4+2z^2(xy+xz+yz)+(xy+xz+yz)^2)\geq0$$ or $$\prod_{cyc}(x-y)\sum_{cyc}(y-x)(z^4+2z^2(xy+xz+yz))\geq0$$ or $$\prod_{cyc}(x-y)\sum_{cyc}(x^4z-x^4y)+2(xy+xz+yz)\prod_{cyc}(x-y)\sum_{cyc}(x^2z-x^2y)\geq0$$ or $$\prod_{cyc}(x-y)^2\sum_{cyc}(x^2+3xy)\geq0.$$