Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$

Question

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 $x$.

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[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

Answer 1

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 $x=y=9$.

Answer 2

Hint

$a+b=27,a,b\ge0$

$a^{2/3}+b^{2/3}=a^{2/3}+(27-a)^{2/3}=f(a)$

Use http://mathworld.wolfram.com/SecondDerivativeTest.html

Answer 3

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.

Answer 4

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 $x_{min} = 9$.