Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$ Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then 
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}=\frac{a}{a+b+\frac{ca}b}$$ which is in general not less than $\frac{a}{a+b+c}$.
So what can I do? Do I have to use some Hölder?

Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, by Muirhead $$\sum_{cyc}\frac{ab}{ab+b^2+ca}=\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^3+(zx)^3}\leq$$
$$\leq\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^2\cdot(zx)+(zx)^2\cdot(y^2)}=\sum_{cyc}\frac{x^2y}{x^2y+y^2z+z^2x}=1.$$
