show that $f(x) = \frac{x^2}{ 1 + \vert x \vert} $ is uniformly continuous on $\mathbb{R}$

Question

I started proving it by saying that on [-1;1] the map is continuous and so we have uniform continuity. Then, I want to show that f is uniform continuous on [1; $\infty$[ and ]-$\infty$ ; -1]. I wanted to do it with the definition, I have the following expression: $\forall x,y \in [1; \infty$[ \ $\mid f(x) - f(y) \mid = \mid$ $\frac{x^2}{1 + \mid x \mid} - \frac{y^2}{ 1 + \mid y \mid} \mid = \mid \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \mid = \mid \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)} \mid$ I don't know how to conclude. I also have to show that the result stays true for $x \in [1; \infty$[ and y $\in ]-\infty ; -1]$ ?

Answer 1

You are almost there as $f$ is even and for $x,y \ge 0$

$$\begin{aligned}\left\vert f(x) - f(y) \right\vert &= \left\vert \frac{x^2}{1 + \vert x \vert} - \frac{y^2}{ 1 + \vert y \vert} \right\vert\\ &= \left\vert \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \right\vert\\ &= \left\vert \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)}\right\vert\\ &\le \vert x - y \vert \end{aligned}$$

knowing that $0 \le x+ y + xy \le (1+x)(1+y)$

Answer 2

First, as the function is even, we may suppose $x,y\ge 0$. In this case, $$|f(x)-f(y)|=|x-y|\frac{x+y+xy}{(1+x)(1+y)}<|x-y|\frac{1+x+y+xy}{(1+x)(1+y)}=|x-y|.$$