Cyclic Inequality $\frac{a^3}{(a+b)(a+c)}$

Question

For positive real numbers $a,b,c$ prove that $$ \sum{\frac{a^3}{(a+b)(a+c)}} \geq \frac{a+b+c}{4}$$

Source: high school olympiad inequality

I didn't succeed in any way, I am quite sure it will be done with multiple AG inequalities

Answer 1

Using AM-GM $$\frac{a^3}{(a+c)(a+b)} + \frac{a+b}{8} + \frac{a+c}{8} \geq \frac{3a}{4}$$

Suming up all 3 variants yields inequality