$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}$

Question

I am reading "Introduction to Set Theory and Topology" (in Japanese) by Kazuo Matsuzaka.

Problem 18 on p.194
Let $f$ be a function from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$ such that $f(x_1,x_2):=\frac{x_1x_2}{x_1^2+x_2^2}$ and $f(0,0):=0$.
Let $c$ be an arbitrary real number.
Then, $x_1\mapsto f(x_1,c)$ and $x_2\mapsto f(c,x_2)$ are continuous functions from $\mathbb{R}$ to $\mathbb{R}$.
Show that $f$ is not a continuous function from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$.

The author's solution:
For example, $$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}$$ and this is not a neighborhood of $(0,0)$.

I filled the gap of this answer as follows:

If $x_1=0$ and $x_2=0$, then $f(x_1,x_2)=0$.
If $x_1=0$ and $x_2\neq 0$, then $f(x_1,x_2)=0$.
If $x_1\neq 0$ and $x_2=0$, then $f(x_1,x_2)=0$.
Let $x_1\neq 0$ and $x_2\neq 0$.
Then $\left|f(x_1,x_2)\right|=\frac{|x_1||x_2|}{|x_1|^2+|x_2|^2}=\frac{1}{\frac{|x_1|}{|x_2|}+\frac{|x_2|}{|x_1|}}$.
By a famous inequality,
$$\frac{\frac{|x_1|}{|x_2|}+\frac{|x_2|}{|x_1|}}{2}\geq\sqrt{\frac{|x_1|}{|x_2|}\cdot\frac{|x_2|}{|x_1|}}=1.$$
So, $$\frac{|x_1|}{|x_2|}+\frac{|x_2|}{|x_1|}\geq 2.$$
So, $$|f(x_1,x_2)|=\frac{1}{\frac{|x_1|}{|x_2|}+\frac{|x_2|}{|x_1|}}\leq\frac{1}{2}.$$
And $|f(x_1,x_2)|=\frac{1}{\frac{|x_1|}{|x_2|}+\frac{|x_2|}{|x_1|}}=\frac{1}{2}$ if and only if $x_1=\pm x_2$.

Therefore, $$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}$$ and this is not a neighborhood of $(0,0)$.

My solution used the famous inequality about arithmetic mean and geometric mean.
But I don't want to use this inequality.
Please show the following equality without using the famous inequality $$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}.$$

Answer 1

Let $g(x) = \frac{x}{1+x^2}$.
Then, $g'(x) = \frac{1-x^2}{(1+x^2)^2}$.
If $-1<x<1$, then $g'(x)>0$.
If $x<-1$ or $1<x$, then $g'(x)<0$.
If $x=1$ or $x=-1$, then $g'(x)=0$.
And $g(1)=\frac{1}{2}$ and $g(-1)=-\frac{1}{2}$.

If $x_1=0$, then $f(x_1,x_2)=0$.
Let $x_1\neq 0$.
Then, $f(x_1,x_2)=\frac{x_1x_2}{x_1^2+x_2^2}=\frac{\frac{x_2}{x_1}}{1+\left(\frac{x_2}{x_1}\right)^2}=g\left(\frac{x_2}{x_1}\right)$.
$|f(x_1,x_2)-f(0,0)|=|f(x_1,x_2)|=\left|g\left(\frac{x_2}{x_1}\right)\right|\leq\frac{1}{2}$.
And $\left|g\left(\frac{x_2}{x_1}\right)\right|=\frac{1}{2}$ if and only if $x_1=x_2$ or $x_1=-x_2$.
So, the following equality holds:
$$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}.$$