I want to prove that $f(x)=\frac{x^2}{x+1}$ is uniformly continuous on $[0, \infty). $
Let $y=x+\delta>x$
$$\implies \lvert f(x)-f(y) \rvert=\lvert f(x)-f(x+\delta) \rvert= \vert{\frac{x^2}{x+1}-\frac{(x+\delta)^2}{x+\delta+1}}\lvert$$
since $\frac{x^2}{x+1}$ is always increasing on $[0, \infty)$ it follows that $f(x+\delta)>f(x)$ $$\implies-\left(-\frac{(x+\delta)^2}{x+\delta+1}+\frac{x^2}{x+1}\right)=\frac{(x+\delta)^2}{x+\delta+1}-\frac{x^2}{x+1} \\<\frac{(x+\delta)^2}{x+1}-\frac{x^2}{x+1} \color{red}{<}\frac{(x+\delta)^2}{\color{red}{x+\delta}}-\frac{x^2}{x+1}$$
since $\frac{x^2}{x+1}<x$
$$\implies \frac{(x+\delta)^2}{x+\delta}-\frac{x^2}{x+1}<x+\delta-x=\delta \\\implies \text{choose } \delta=\epsilon \\ \implies\lvert f(x)-f(y)\rvert<\epsilon$$
The $\color{red}{\text{red}}$ part ist what I am unsure about. This inequality basically only works when $\delta<1$ and I am not sure if it must be true for all deltas? (I am not really looking for alternate ways to prove it because I kind of want to figure it out by myself. I am only unsure about the red part. If it is wrong and my proof doesn't make sense I'll just give it another go)