inequality for positive real numbers Given $$a,b,c,x,y,z \in R^+$$ how to show the following inequality:
$$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \geq \frac{(a+b+c)^3}{3(x + y + z)}$$
I rearranged the inequality since all are positive, the inequality would be true iff
$$\frac{a^3}{x(a+b+c)^3}+\frac{b^3}{y(a+b+c)^3}+\frac{c^3}{z(a+b+c)^3}\geq\frac{1}{3(x+y+z)}$$
which further simplifies to
$$\frac{a^3yz}{(a+b+c)^3}+\frac{b^3xz}{(a+b+c)^3}+\frac{c^3xy}{(a+b+c)^3}\geq\frac{xyz}{3(x+y+z)}$$
and now I am struggling how to rearrange further to apply Holding's inequality

By Holder $$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}=\frac{(1+1+1)\left(\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}\right)(x+y+z)}{3(x+y+z)}\geq\frac{(a+b+c)^3}{3(x+y+z)}.$$
