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)

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 $m = n = p = 1 \implies x = y = z = 1$.
