<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] 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: $8x^4 = \dfrac{81}{2}, 27y^4 = 27, 64z^4 = \dfrac{81}{4}\implies x = \dfrac{3}{2}, y = 1, z = \dfrac{3}{4}$
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