inequality for positive real numbers

Question

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

Answer 1

Apply Sedakryan's inequality https://en.m.wikipedia.org/wiki/Sedrakyan%27s_inequality $\\$ For $n=3$ , $\small{\frac{a_{1}^2 }{b_1 } +\frac{a_{2}^2 }{b_2 } +\frac{a_{3}^2 }{b_3 } \ge \frac{( a_1 +a_2 +a_3 )^2 }{b_1 + b_2 +b_3 }}$.

$\\$ Set $a_1 =a^{3/2} ,a_2 =b^{3/2} , a_3 =c^{3/2} , b_1=x ,b_2 =y , b_3 =z$.

$\\$ The inequality must be simplified like $a^3 /x +b^3 /y +c^3 /z \ge \frac{(a^{3/2} +b^{3/2} +c^{3/2 } )^2 }{(x+y+z)}$.

$\\$ Apply $ (a^{3/2} + b^{3/2}+c^{3/2} )^2 \ge \frac{1}{3} (a+b+c)^3 $ to complete the proof.

Answer 2

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