<s>[INST] Proving the inequality $\arctan\frac{\pi}{2}\ge1$ Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.
 [/INST] By  Taylor's theorem with remainder in the integral form for the function $f=\arctan$ at the point $1$, we have:
\begin{equation}\arctan x=\arctan 1+(x-1)f'(1)+\int_1^x\frac{f''(t)}{2!}(x-t)dt \end{equation}
then we substitute $\displaystyle f'(1)=\frac{1}{2}$, $\displaystyle f''(t)=\frac{-2t}{(1+t^2)^2}$ and $\displaystyle x=\frac{\pi}{2}$  we find
\begin{equation}\arctan \frac{\pi}{2}=\frac{\pi}{4}+(\frac{\pi}{2}-1)\frac{1}{2}-\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt\\=1+\frac{\pi-3}{2} -\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt.\end{equation}
Now, we have
\begin{align}D=\frac{\pi-3}{2} -\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt&\geq \frac{\pi-3}{2} -(\frac{\pi}{2}-1)\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}dt\\&=\frac{\pi-3}{2}-(\frac{\pi}{2}-1)\frac{1}{4}\frac{\pi^2-4}{\pi^2+4}=S.\end{align}
The approximation $\pi\approx 3.1416$ gives $D\geq S\approx 4.27\ 10^{-3}$ so we conclude: 
$$\arctan\frac{\pi}{2}\geq 1.$$
</s>