Solve the inequality $ \sum_{cyc}\frac{a-bc}{a+bc} \le \frac32$

Question

Solve the inequality $ \displaystyle\sum_{cyc}\frac{a-bc}{a+bc} \le \frac32$ given $a + b + c = 1$ and $a, b, c \in \mathbb{R_{>0}}$

So, I wanted to use the known inequality $9(a+b)(b+c)(c+a) \ge 8(a+b+c)(ab+bc+ca)$.

In order to do so, I tried some algebraic manipulations and reduced the expression $\displaystyle\sum_{cyc}\frac{a-bc}{a+bc}$ to $\frac{(ab+bc+ca)+3abc}{(a+b)(b+c)(c+a)}$ by using the fact that $a+b+c=1$

Now I am stuck and I am not able to proceed further. Can someone please help me.

Answer 1

Thank you @projectilemotion

So I proceeded as follows :

$\sum_{cyc} \frac{a-bc}{a+bc}\le \frac{3}{2}$

$\iff\sum_{cyc} \frac{a-bc}{a+bc} +3\le \frac{3}{2}+3$

$\iff\sum_{cyc} (\frac{a-bc}{a+bc}+1)\le \frac{9}{2}$

$\iff \sum_{cyc} \frac{2a}{a+bc}\le \frac{9}{2}$

However, $a+bc=1-b-c+bc=(1-b)(1-c)=(a+b)(a+c)$

So the inequality becomes $\sum \frac{2a}{(a+b)(a+c)}\le \frac{9}{2}$

$\iff \frac{\sum_{cyc} 2a(b+c)}{(a+b)(b+c)(c+a)}\le \frac{9}{2}$

$\iff 8(ab+bc+ca)\le 9(a+b)(b+c)(c+a)$

$\iff 8(a+b+c)(ab+bc+ca)\le 9(a+b)(b+c)(c+a)$

which is true. See here.