<s>[INST] 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
 [/INST] Let $\frac{x-y}{2}=a$ and $\frac{x+y}{2}=b$.
Thus, $a\geq0$, $b\geq0$, $x=a+b$ and $$\sqrt{a^3}+\sqrt{b^3}=27.$$
Now, let $f(x)=\sqrt{x^3}.$ 
Thus, $f$ is a convex function on $[0,+\infty)$ and by Karamata $$27=\sqrt{a^3}+\sqrt{b^3}\leq\sqrt{(a+b)^3}+\sqrt{0^3},$$ which gives
$$x=a+b\geq9.$$
The equality occurs for $b=0$, which says that $9$ is a minimal value.
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