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?

Define:
$$u^4=8x^4,~v^4=27y^4,~w=64z^4$$
$$f=u^4+v^4+w^4$$
Constraint becomes: $$8^{-1/4}u+27^{-1/4}v+64^{-1/4}w=13/4$$
So we have:
$$a=8^{-1/4},~~b=27^{-1/4},~~c=64^{-1/4}$$
Now you can use C-S inequality.
