pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$

I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?

Let $\,u = 2x, v=3y, w=4z\,$ then the problem is to minimize $u^4 / 2 + v^4 / 3 + w^4 / 4$ under the constraint $u / 2 + v / 3 + w / 4 = 13/4$. By the weighted power means inequality:
$$
\sqrt[4]{\frac{u^4 / 2 + v^4 / 3 + w^4 / 4}{1/2+1/3+1/4}} \;\ge\; \frac{|u| / 2 + |v| / 3 + |w| / 4}{1/2+1/3+1/4} \;\ge\; \frac{13/4}{13/12} = 3
$$
The minimum is attained for $\,u=v=w=3\,$.
