<s>[INST] 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?
 [/INST] Let $x=\dfrac{b}{a},y=\dfrac{c}{b},z=\dfrac{a}{c},\Longrightarrow xyz=1$
then
$$\dfrac{a^3}{a^3+b^3+abc}=\dfrac{1}{1+x^3+\frac{x}{z}}=\dfrac{1}{xyz+x^3+x^2y}=\dfrac{xyz}{xyz+x^3+x^2y}=\dfrac{yz}{yz+x^2+xy}$$
and use Cauchy-Schwarz inequality we have
$$\left(\sum_{cyc}\dfrac{yz}{yz+x^2+xy}\right)\sum_{cyc}yz(yz+x^2+xy)\ge (xy+yz+xz)^2$$
This is clear 
In fact
$$(xy+yz+xz)^2=yz(yz+x^2+xy)+zx(zx+y^2+yz)+xy(xy+z^2+zx)$$
so
$$\sum_{cyc}\dfrac{yz}{yz+x^2+xy}\ge 1$$
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