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.

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