Question: Let $E=\{(x,y)\in \Bbb R^2 \mid x^2+y^2 \leq 1, \ x \geq 0, \ y\geq0\}$. Let $m(x,y) = \displaystyle\frac{2}{\frac 1x +\frac 1y}$. Find $\max_Em(x)$ if exists.
Background: This question has been asked to me at my real analysis exam. I don't know anything about finding maximum or minimum of a two variable function so I'm trying to solve it only with my knowledge.
Answer The first thing I can observe is that $E$ is a compact set. Unfortunately $m(x,y)$ is not continuous on E since is not defined in $0$ so I can not apply the Weierstrass theorem. If I keep fixed $x$ and let $y \to 0$ I observe that $m(x,y) \to 0$. Vice versa if I keep $y$ fixed and let $x \to 0$ also $m(x,y) \to 0$. At this point (I'm not sure about this) I can say that $$ \tilde m(x,y) \doteqdot \begin{cases} m(x,y) & \text{if $x \neq 0$ and $y\neq0$} \\ 0 & \text{if $x = 0$ or $y=0$} \end{cases}$$ $\tilde m(x,y)$ is continuous on $E$ so it must have maximum for the Weierstrass theorem. Since $m(x,y) >0$ in all of his domain maximum can not be 0. At this point looking at the equation I come up with the idea that the maximum is in $\left(\displaystyle\frac{\sqrt{2}}2,\frac{\sqrt{2}}2\right)$ and this is my proof. I need to demonstrate that $$m\left(\displaystyle\frac{\sqrt{2}}2,\frac{\sqrt{2}}2\right) \geq m(x,y) \quad \forall \ x,y \in E \ \text{with} \ x \neq 0 \ \text{and} \ y\neq0$$
$$\frac{\sqrt2}2 \geq^{?} \frac{2xy}{x+y}$$
$$\sqrt2(x+y) \geq^{?} 4xy$$
$$x+y-2\sqrt2xy \geq^{?} 0$$
At this point using the AM-GM inequality I obtain that:
$$\sqrt{\mathstrut xy}-\sqrt{\mathstrut 2}xy \geq 0 \implies x+y-2\sqrt{\mathstrut 2}xy \geq 0$$
so I can study the first inequality that become:
$$\sqrt{\mathstrut 2xy} -1 \leq 0$$
Since $y \leq \sqrt{\mathstrut 1-x^2}$ I have that:
$$\sqrt[4]{4x^2-4x^4}-1 \leq 0 \implies \sqrt{\mathstrut 2xy} -1 \leq 0$$
Let $f(x)= \sqrt[4]{4x^2-4x^4}-1$. I just need to find the maximum of $f(x)$ and see that it is less equal than 0.
$$f'(x)= \frac{2x-4x^3}{\sqrt[4]{(4x^2-4x^4)^3}}$$
$f(x)$ has maximum in $\displaystyle\frac{\sqrt2}2$ and I have that $f\left(\displaystyle\frac{\sqrt2}2\right) = 0 \leq 0$
Finally I can say that $m(x,y)$ has maximum in $\left(\displaystyle\frac{\sqrt{2}}2,\frac{\sqrt{2}}2\right)$
My questions are:
- Is my proof correct and formal?
- Are there any better method to prove my conjecture with my knowledge?
Thank you very much!
I answer to (2).
$m(x,y)$ is the harmonic mean between $x$ and $y$, which is at most the quadratic mean $$ \sqrt{\frac{x^2+y^2}{2}}=\frac{\sqrt{2}}{2} $$ with equality iff $x=y=\sqrt{2}/2$.