I have proven that (and we are required to use this) $y^2\arctan y-x^2\arctan x\geq(y^2-x^2)\arctan x$ (the proof was near-trivial). I now have to use this to show that $f(x)=x^2\arctan x$ is not uniformly continuous in $\Bbb R$.
Here is what I have so far (I'm not used to writing mathematical proofs in English so bear with me):
Let there be $\delta>0$. Let $x=\frac{10}{\delta}, y=x+\frac{\delta}{2}$. It follows that $|x-y|<\delta$, and:
$\begin{align}|f(x)-f(y)| &=|x^2\arctan x-y^2\arctan y|\\ &\geq(y^2-x^2)\arctan x\\ &=\left(\left(\frac{10}{\delta}+\frac{\delta}{2}\right)^2-\left(\frac{10}{\delta}\right)^2\right)\arctan\frac{10}{\delta}\\ &=\left(\frac{100}{\delta^2}-\frac{100}{\delta^2}+2\frac{10\delta}{2\delta}+\frac{\delta^2}{4}\right)\arctan\frac{10}{\delta}\\ &=\left(10+\frac{\delta^2}{4}\right)\arctan\frac{10}{\delta}\end{align}$
And this is where I realized that my choice of $x$ and $y$ was probably not wise - I have failed to prove that the function is greater than some concrete number, as I can make no assumption about the value of $\arctan\frac{10}{\delta}$. I have failed to find better variables - I'd appreciate any help/guidance.