To show: the uniform continuity of $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{1}{1+x^4}$ per Lipschitz-continuity or $\varepsilon - \delta$-proof.
What I tried:
Let $x, y \in \mathbb{R}$ with $|x-y| \lt \delta$, it follows:
$|f(x)-f(y)| \lt \varepsilon \Leftrightarrow |\frac{1}{1+x^4} - \frac{1}{1+y^4}| \lt \varepsilon$. I then tried to reaarrange the term so that I would see the fitting $\delta$ or the needed estimation for the Lipschitz-continuity, but it didn't work out.
Can someone help at this? Thanks in advance!
Note that $|x|^{3} \leq 1+x^{4}$. This is because $|x| \leq 1$ implies $|x|^{3} \leq 1 \leq 1+x^{4}$ and $|x| >1$ implies $|x|^{3} \leq x^{4} \leq 1+x^{4}$.
Hence $|f'(x)|=|\frac {-4x^{3}} {(1+x^{4})^{2}}|\leq 4$. By MVT we get $|f(x)-f(y) | \leq 4|x-y|$ for all $x,y$.