Prove this inequality $\sum_{cyc}\frac{ab}{c\left(c+a\right)}\ge \sum_{cyc}\frac{a}{a+c}$

Question

Given $a,b,c>0$.Prove that: $$\frac{ab}{c\left(c+a\right)}+\frac{bc}{a\left(a+b\right)}+\frac{ca}{b\left(b+c\right)}\ge \frac{a}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}$$

---

$\Leftrightarrow \frac{a^4c^2-a^3b^3-a^3bc^2+a^3c^3+a^2b^4-a^2b^3c-3a^2b^2c^2-ab^2c^3+b^3c^3+b^2c^4}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}\ge 0$

Help me to countinue or give me some suggestion. Thanks

Answer 1

we have $$a^3b^3+a^3c^3+b^3c^3\geq 3a^2b^2c^2$$ by $AM-GM$ then we have to Show that $$a^4c^2+a^2b^4+b^2c^4\geq a^3bc^2+a^2b^3c+ab^2c^3$$ let $$a\geq b\geq c$$ then we have $$c^2(a-b)(a^3-b^2c)+(b-c)b^2(a^2b-c^3)\geq 0$$ we can prove this inequality (the last one) without the assumption $$a\geq b\geq c$$ we use that $$x^2+y^2+z^2\geq xy+yz+zx$$ writing $$(a^2c)^2+(ab^2)^2+(bc^2)^2\geq a^3b^2c+a^2bc^3+ab^3c^2$$ thank you for your comment dear Michael!

Answer 2

We need to prove that: $$\sum_{cyc}(a^4c^2+a^3b^3-a^3c^2b-a^2b^2c^2)\geq0,$$ which is true because $\sum\limits_{cyc}a^4c^2\geq3a^2b^2c^2$ by AM-GM and $$\sum_{cyc}a^3b^3\geq \sum_{cyc}a^3c^2b$$ is true by Rearrangement.

Indeed, we need to prove that $$\sum_{cyc}(ab)^3\geq\sum_{cyc}(ac)^2(ab).$$