Inequality Challenge: Prove Real $a,b,c,x,y,z$

[anemone]

Prove for all positive real $a,\,b,\,c,\,x,\,y,\,z$ that $\dfrac{a^3}{x}+\dfrac{b^3}{y}+\dfrac{c^3}{z}\ge \dfrac{(a+b+c)^3}{3(x+y+z)}$.

 

[anemone]

Hint:

Spoiler

Use the Cauchy-Schwarz inequality twice and then Holder's inequality once will be sufficient...

 

[anemone]

My solution:

Spoiler

Use the Cauchy Schwarz inequality, the LHS of the given inequality becomes greater than:

$\dfrac{a^3}{x}+\dfrac{b^3}{y}+\dfrac{c^3}{z}\ge \dfrac{(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}})^2}{x+y+z}$

Next, impose the Holder's inequality on $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}$, we see that we have:

$a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\ge\sqrt{a^2+b^2+c^2}\sqrt{a+b+c}$

Thus $(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}})^2\ge (a^2+b^2+c^2)(a+b+c)$.

Use the Cauchy Schwarz inequality again we have $3(a^2+b^2+c^2)≥(a+b+c)^2$.

At last, the combined result leads us to the desired proof:

$\begin{align*}\dfrac{a^3}{x}+\dfrac{b^3}{y}+\dfrac{c^3}{z}&\ge \dfrac{(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}})^2}{x+y+z}\\&\ge \dfrac{(a^2+b^2+c^2)(a+b+c)}{x+y+z}\\&\ge \dfrac{(a+b+c)^2(a+b+c)}{3(x+y+z)}\\&\ge \dfrac{(a+b+c)^3}{3(x+y+z)}\,\,\,\,\text{Q.E.D.}\end{align*}$