<s>[INST] 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?
 [/INST] Using Lagrange multiplier method:
$$\frac{\partial}{\partial x}\left(8x^4+27y^4+64z^4-\lambda(x+y+z-\frac{13}4)\right)=0\\4\cdot8x^3=\lambda$$
Similarly for $y$ and $z$ you get:
$$4\cdot 27y^3=\lambda\\4\cdot 64z^3=\lambda$$
If we use $$\frac\lambda4=\mu^3$$
one can take the cubic root and get $$x=\frac\mu2\\y=\frac\mu3\\z=\frac\mu 4$$
Plugging it into the constraint:
$$\frac\mu2+\frac\mu3+\frac\mu4=\frac{13}{4}\\\mu\frac{6+
4+3}{12}=\mu\frac{13}{12}=\frac{13}4$$
So $\mu=3$. Then $$x=\frac32\\y=1\\z=\frac34$$
The minimum value of the function is then $$8\frac{3^4}{2^4}+27+64\frac{3^4}{4^4}=\frac{351}4$$
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