If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$

Question

If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$

I have tried the following,in decreasing order of success:

1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$

2)Substituting $1=abc$ yields nothing

3)Substituting $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ yields something weird

4)Rearrangement inequality on the sequences $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}$

5)Titu's lemma doesn't help

A little nudge in the right direction would help.

Answer 1

(I) By rearrangement inequality:

$\displaystyle \begin{align} &\sum_{cyc} \frac{1}{b(a+b)} \ge \sum\limits_{cyc} \frac{1}{b(a+c)} \tag{1} \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{b(a+c)}\right) = \sum_{cyc} \frac{1}{2}\left(\frac{1}{b(a+b)} + \frac{1}{c(a+b)}\right) \\ \iff & \sum_{cyc} \frac{1}{b(a+b)} \ge \frac{1}{2}\sum_{cyc} \frac{b+c}{bc(a+b)} \end{align}$

By Am-Gm Inequality :

$$\sum_{cyc} \frac{b+c}{bc(a+b)} \ge 3\sqrt[3]{\prod\limits_{cyc} \frac{b+c}{bc(a+b)}} = 3$$

Thus establishing desired inequality.

Note: $(1)$ can be viewed as a consequence of CS as well.

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

(II) Substituting $\displaystyle a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$:

The inequality required to prove becomes:

$$\sum\limits_{cyc} \frac{x^2}{z^2+xy} \ge \frac{3}{2}$$

We may rewrite LHS as $\displaystyle \sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y}$ and apply Cauchy-Schwarz Inequality:

$$\sum\limits_{cyc} \frac{x^4}{x^2z^2+x^3y} \ge \frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc} x^2z^2 + \sum\limits_{cyc} x^3y}$$

So it suffices to prove that: $$2(x^2+y^2+z^2)^2 \ge 3\sum\limits_{cyc} x^2z^2 + 3\sum\limits_{cyc} x^3y \\ \iff (x^4+y^4+z^4) + \sum\limits_{cyc} (x^4 + x^2y^2) \ge 3\sum\limits_{cyc} x^3y$$

This is the consequence of adding the following:

(i) The Rearrangement Inequality: $x^4+y^4+z^4 \ge x^3y+y^3z+z^3x$

(ii) The Am-Gm Inequality: $\displaystyle \sum\limits_{cyc} (x^4 + x^2y^2) \ge 2\sum\limits_{cyc} \sqrt{x^6y^2} = 2\sum\limits_{cyc} x^3y$

Answer 2

Hint:

Indeed, as @Macavity pointed out, the sum $s(a,b,c)$ only has circular symmetry. In fact it is not hard to show ( the same method as below) that if $a\le b\le c$ then $s(a,b,c)\le s(c,b,a)$.

Here is a method amenable to a computer algebra system. As suggested, substitute $a=u/v$, $b=v/w$, $c=w/u$. Get an equivalent inequality homogeneous of degree $6$ in $u$,$v$, $w\ge 0$

$$2 u^4 v^2-3 u^3 v^3+2 u v^5+2 u^5 w-3 u^4 v w+4 u^2 v^3 w- 3 u v^4 w+4 u^3 v w^2-\\-6 u^2 v^2 w^2+2 v^4 w^2-3 u^3 w^3+4 u v^2 w^3-3 v^3 w^3+2 u^2 w^4-3 u v w^4+2 v w^5\ge 0$$

with circular symmetry. It is enough to consider the cases $u\le v \le w$ and $u \ge v \ge w$.

In the first case, make the substitution $u=p$, $v=p+q$, $w=p+q+r$, with $p$, $q$, $r\ge 0$. One gets $$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+31 p^3 q^2 r +38 p^2 q^3 r+22 p q^4 r+\\+5 q^5 r+8 p^4 r^2+41 p^3 q r^2 +66 p^2 q^2 r^2+46 p q^3 r^2+13 q^4 r^2+14 p^3 r^3+43 p^2 q r^3+\\ +43 p q^2 r^3+17 q^3 r^3+9 p^2 r^4+17 p q r^4+10 q^2 r^4+2 p r^5+2 q r^5$$

clearly positive, with equality when $q=r=0$, that is, the $a$,$b$,$c$ are equal. The other case is treated similarly, giving the positive expression $$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+23 p^3 q^2 r+22 p^2 q^3 r+8 p q^4 r+q^5 r+8 p^4 r^2+33 p^3 q r^2+42 p^2 q^2 r^2+18 p q^3 r^2+3 q^4 r^2+14 p^3 r^3+35 p^2 q r^3+23 p q^2 r^3+5 q^3 r^3+9 p^2 r^4+11 p q r^4+2 q^2 r^4+2 p r^5$$

Answer 3

Look at $\frac{1}{b(a+b)}$ and $\frac{1}{a(a+b)}$. Adding together yields $\frac{1}{ab}$. So the sum is just $$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{a + b + c}{abc}$$ The denominator is 1 and the numerator is at least 3 by AM-GM for positive $a,b,c$.

Answer 4

I think the inequality is false. Currently I am searching for a triplet which disproves it. But here is my argument- Substitute $a=x/y....$. Now write your expression in terms of $x,y,z$ and then apply Titu's lemma. You should get $$\sum_{cyc}\frac{x^2}{xy+z^2}\geq\frac{(x+y+z)^2}{x^2+y^2+z^2+xy+yz+zx}.$$ This means after opening it and making it greater than $1/2$, you get a positive quantity less than $0$. Not possible right. Please tell me if I am wrong anywhere and also please edit this answer(I don't know what is wrong in my syntax).