How to prove $\left|x^2\arctan(x)-y^2\arctan(y)\right|\leqslant\left(1+\frac\pi2\right)|x-y|\;?$

For any $x,y\in[0,1]$ prove that

$\left|x^2\arctan(x)-y^2\arctan(y)\right|\leqslant\left(1+\dfrac\pi2\right)|x-y|\;.\\$

Proof:

$f(x)=x^2\arctan(x)$ is differentiable on $(0,1)$ and continuous on $[0,1]$, then by the mean value theorem there exists a point $c$ between $x$ and $y$$\big(\!\Rightarrow c\in(0,1)\big)$ such that

$\dfrac{|x^2\arctan(x)-y^2\arctan(y)|}{|x-y|}=|f′(c)|\;.$

Since $\;f′(c)=2c\arctan(c)+\dfrac{c^2}{c^2+1}\;,\;$ it follows that

$\dfrac{|x^2\arctan(x)-y^2\arctan(y)|}{|x-y|}=\left|2c\arctan(c)+\dfrac{c^2}{c^2+1}\right|$

$\leqslant\left|2c\arctan(c)\right|+\left|\dfrac{c^2}{c^2+1}\right|$

$\leqslant2c\dfrac\pi4+1\leqslant\dfrac\pi2+1\,.$

Did I make a mistake? Because I don’t know how to continue from here.