contest math - Inequality, Cauchy Schwarz and Schur

For $a,b, c>0$, prove that

$$\frac{a^3}{a^3+b^3+abc}+\frac{b^3}{b^3+c^3+abc}+\frac{c^3}{c^3+a^3+abc}\geq 1$$

I tried the following

$$\sum_{cyc}\frac{a^3}{a^3+b^3+abc}\cdot \sum_{cyc}a^3+b^3+abc\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
where cyc stands for cyclic sum of $a,b,c$, then we could show that $$(a^{3/2}+b^{3/2}+c^{3/2})^2\geq 2(a^3+b^3+c^3)+3abc$$ to prove our original inequality. This looks similar to Schur but with inequality sign reversed. How would you proceed? Any ideas/hint?