Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$ Let $x,y\in \mathbb{R}$ such that $$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$. Find the minimize value of $x$.

[Edit by Michael Rozenberg] I tried to use AM-GM to retire the radical but failed:
$$27=\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\geq2\sqrt[4]{\left(\frac{x+y}{2}\right)^3\left(\frac{x-y}{2}\right)^3}=$$
$$=2\sqrt[4]{\frac{(x^2-y^2)^3}{64}}=\sqrt[4]{\frac{(x^2-y^2)^3}{4}}.$$
 Help me

Note that $x(y)$ is an even function of $y$, with the domain $y\le |x|$. So, just examine $0\le y \le x $. Evaluate,
$$x'(y) =\frac{-y}{\left(\sqrt{\frac{x+y}{2}} +\sqrt{\frac{x-y}{2}}\right)^2}<0$$
i.e. $x(y)$ strictly decreases for $0< y \le x $. Therefore, the minimum is at $y=x$, and by symmetry also at $y=-x$. Plug $y=\pm x$ into
$$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$
to obtained $x_{min} = 9$.
