let $x,y,z$ be positive numbers,and such $x+y+z=3$,show that $$\dfrac{x}{x^3+y^2+z}+\dfrac{y}{y^3+z^2+x}+\dfrac{z}{z^3+x^2+y}\le 1$$
My try:$$(x^3+y^2+z)(\dfrac{1}{x}+1+z)\ge 9$$ so $$\dfrac{x}{x^3+y^2+z}+\dfrac{y}{y^3+z^2+x}+\dfrac{z}{z^3+x^2+y}\le\dfrac{6+xy+yz+xz}{9}\le 1$$
Have other nice methods? Thank you
Note that $2+x^3=x^3+1+1\geqslant 3x$, $y^2+1\geqslant 2y$, thus$$\dfrac{x}{x^3+y^2+z}=\frac{x}{3+x^3+y^2-x-y}\leqslant\frac{x}{3x+2y-x-y}=\frac{x}{2x+y}.$$Similarly, we can get $$\dfrac{y}{y^3+z^2+x}\leqslant\frac{y}{2y+z},\,\,\,\,\,\dfrac{z}{z^3+x^2+y}\leqslant\frac{z}{2z+x}.$$It suffices to show$$\frac{x}{2x+y}+\frac{y}{2y+z}+\frac{z}{2z+x}\leqslant 1\iff \frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\geqslant 1.$$By Cauchy's inequality, we get$$(y(2x+y)+z(2y+z)+x(2z+x))\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)\geqslant (x+y+z)^2.$$As $y(2x+y)+z(2y+z)+x(2z+x)=(x+y+z)^2$, so $$\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\geqslant 1.$$
