Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$

@Michael Rosenberger Thanks, here is my $0.02:

Write the whole inequality in terms of $ab, ac, bc$:

$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{ac}{ac+bc}+\frac{ab}{ab+ac}+\frac{bc}{bc+ab}$$

Then use the following inequality proved here - as a Lemma inside Andreas's answer:

$$ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$

for $a\leftarrow ac$, $b\leftarrow ab$, $c\leftarrow bc$

That's all.

This proof I found a few minutes ago.

I hope that there is something nicer.

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

Hence, we need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)\geq\frac{w^2}{2v^2}-\frac{1}{2}$$ or $$\sum\limits_{cyc}(a-b)(a+c)(b+c)\geq\left(\frac{w^2}{v^2}-1\right)\prod_{cyc}(a+b)$$ or $$\sum\limits_{cyc}(a-b)c^2\geq\left(\frac{w^2}{v^2}-1\right)(9uv^2-w^3)$$ or $$(9uv^2-w^3)(v^2-w^2)\geq v^2(a-b)(b-c)(c-a)$$ and since $v^2\geq w^2$, it remains to prove that $$(9uv^2-w^3)^2(v^2-w^2)^2\geq v^4(a-b)^2(b-c)^2(c-a)^2$$ or $$(9uv^2-w^3)^2(v^2-w^2)^2\geq 27v^4(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $f(u)\geq0$, where $$f(u)=108v^4w^3u^3-81v^4w^2(2v^2-w^2)u^2-18v^2w^3(10v^4-2v^2w^2+w^4)u+$$ $$+108v^{10}+28v^4w^6-2v^2w^8+w^{10},$$ which says that for $u_{min}$ we get the following equation: $$324u^2v^4w^3-162v^4w^2(2v^2-w^2)u-18v^2w^3(10v^4-2v^2w^2+w^4)=0,$$ which gives $$u_{min}=\frac{6v^3-3vw^2+\sqrt{36v^6+44v^4w^2-7v^2w^4+8w^6}}{12vw}$$.

Thus, it remains to prove that $f\left(u_{min}\right)\geq0$, which gives $$648x^5-396x^4+342x^3+107x^2+20x+8\geq\sqrt{x(36x^3+44x^2-7x+8)^3}$$ or $$(x-1)^2(5832x^8+972x^7+2646x^6+1477x^5+581x^4+105x^3+51x^2-x+1)\geq0.$$ Done!