<s>[INST] find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2}{3(yx^3+zy^3+xz^3)+x^3+y^3+z^3}=\dfrac{3}{yx^3+zy^3+xz^3+1}$ and I don't know what to do next (I think my approach is not right)
 [/INST] Note: The following answer is incorrect, see comments. I don't have a correct version yet. However since the other answer mentions this, I've marked it as community wiki and made it visible.
Let
$$
L
=
\lambda (x^3+y^3+z^3-3)+{x^3\over 3y+1}+{y^3\over 3z+1} +{z^3\over 3x+1}\\
$$
Differentiating with respect to $\lambda, x, y$, and $z$ gives
$$
{\partial L\over\partial\lambda}=x^3+y^3+z^3-3
$$
and three symmetric equations of the form
$$
{\partial L\over\partial x}=3x^2(\lambda+{1\over 3y+1})
$$
Setting $\partial L/\partial\lambda=\partial L/\partial x=...=0$ gives $\lambda=-1/(3y+1)=-1/(3z+1)=-1/(3x+1)$,  so $x=y=z$ and since $x^3+y^3+z^3=3$, $x=y=z=1$.
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