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?
 A: 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 $\,u=v=w=3\,$.
A: 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 $$8\frac{3^4}{2^4}+27+64\frac{3^4}{4^4}=\frac{351}4$$
A: Apply AM-GM inequality repeatedly three times:
$8x^4+\dfrac{243}{2}=8x^4 + \dfrac{81}{2}+\dfrac{81}{2}+\dfrac{81}{2} \ge 4\sqrt[4]{8x^4\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}}=4\sqrt[4]{(27x)^4}= 4|27x|\ge 27(4x)$.
$27y^4 + 81 = 27y^4+ 27+27+27 \ge 4\sqrt[4]{27y^4\cdot 27\cdot 27\cdot 27}=4|27y| \ge 27(4y)$.
$64z^4 + \dfrac{243}{4} = 64z^4 + \dfrac{81}{4}+\dfrac{81}{4}+\dfrac{81}{4} \ge 4\sqrt[4]{64z^4\cdot \dfrac{81}{4}\cdot \dfrac{81}{4}\cdot \dfrac{81}{4}}=4|27z| \ge 27(4z)$.
Now add these three inequalities side by side, we obtain:
$8x^4+27y^4+64z^4 + \dfrac{81\cdot 13}{4} \ge 27(4x+4y+4z) = 27\cdot 13\implies 8x^4+27y^4+64z^4 \ge \dfrac{27\cdot 13}{4}=\dfrac{351}{4}$. This is also the minimum value of the expression $8x^4+27y^4+64z^4$. Equality occurs when: $8x^4 = \dfrac{81}{2}, 27y^4 = 27, 64z^4 = \dfrac{81}{4}\implies x = \dfrac{3}{2}, y = 1, z = \dfrac{3}{4}$
A: Hint:  You can also use Hölder's Inequality, like so:
$$\left(8x^4+27y^4+64z^4 \right) \cdot \left(\frac12+\frac13+\frac14 \right)^3 \geqslant \left(|x|+|y|+|z|\right)^4 \geqslant (x+y+z)^4$$
A: 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.
