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)
 A: The key observation is the term $3x,3y,3z$ at the denominators. This suggests the application of the AM-GM inequality somehow. Indeed, by AM-GM inequality: for $a \ge 0 \implies a^3 + 2 = a^3 + 1 +1 \ge 3a\implies a^3+ 3 \ge 3a+1$. Using this fact for $a = x, y, z$ in the denominators of $A$. Thus: $A \ge \dfrac{x^3}{y^3+3} + \dfrac{y^3}{z^3+3}+\dfrac{z^3}{x^3+3}$. Now let $m = x^3, n = y^3, p = z^3 \implies A \ge \dfrac{m}{n+3}+\dfrac{n}{p+3}+\dfrac{p}{m+3}$, with $m+n+p=3$ and $m,n,p \ge 0$. Using Cauchy-Schwarz inequality twice: $\displaystyle \sum_{\text{cyclic}}\dfrac{m}{n+3} = \displaystyle \sum_{\text{cyclic}}\dfrac{m^2}{mn+3m}\ge \dfrac{3(m+n+p)^2}{3(mn+np+pm)+9(m+n+p)}\ge \dfrac{3\cdot3^2}{(m+n+p)^2+9\cdot 3}=\dfrac{27}{9+27}=\dfrac{27}{36}= \dfrac{3}{4}$. Equality occurs in the original inequality when $m = n = p = 1 \implies x = y = z = 1$.
A: @SuzuHirose's use of standard Lagrange methods indeed finds an extremum.  Here's a color-coded plot of the target function on the cubic surface to show that:

The minima occur when $x = y = z = 1$, etc., as seen in the figure, where blue is small.
A: 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 $x^3+y^3+z^3=3$, $x=y=z=1$.
