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 $x_{min} = 9$.
For any two real numbers $a$ and $b$, we have
$$(a^2+b^2)^3 \geq (a^3+b^3)^2$$
because
$$(a^2+b^2)^3 - (a^3+b^3)^2 = a^2b^2[2a^2+2b^2+(a-b)^2]\geq 0$$
Setting $a=\sqrt{\frac{x+y}{2}}$ and $b=\sqrt{\frac{x-y}{2}}$, we get:
$$\left[\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\right]^2 \leq \left(\frac{x+y}{2}+\frac{x-y}{2}\right)^3=x^3$$
Therefore $x^3 \geq 27^2\Rightarrow x\geq 9$. This minimum is attained when $x=y=9$.
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.
