Proving the inequality $\arctan\frac{\pi}{2}\ge1$ Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.
 A: 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: 
$$\arctan\frac{\pi}{2}\geq 1.$$
A: I don't know what qualifies as nice, but since $\arctan(x) + \arctan(1/x) = {\pi \over 2}$, what you're trying to show is the same as
$$\arctan\bigg({2 \over \pi}\bigg) < {\pi \over 2} - 1$$
Since ${2 \over \pi} < 1$, one has the formula
$$\arctan\bigg({2 \over \pi}\bigg) = \sum_{n=1}^{\infty} {(-1)^{n+1} \over 2n + 1}\bigg({2 \over \pi}\bigg)^{2n + 1}$$
Whenever you have an alternating sum whose terms decrease in magnitude the end result is less than what you get after stopping after a positive term. If we stop at $n = 4$ we get
$$\bigg({2 \over \pi}\bigg) -\bigg({2 \over \pi}\bigg)^3{1 \over 3} + \bigg({2 \over \pi}\bigg)^5{1 \over 5} - \bigg({2 \over \pi}\bigg)^7{1 \over 7} + \bigg({2 \over \pi}\bigg)^9{1 \over 9} = 0.56738523$$
This is less than ${\pi \over 2} - 1 = 0.5707963$.
A: Edited. 
Put $$y:=\sqrt{3}-{\pi\over2}<{5\over31}\ ,$$
where the numerical value has been obtained using continued fractions. Then
$$\arctan\left({\pi\over2}\right)=\arctan(\sqrt{3}-y)=\arctan\sqrt{3}-y\arctan'(\sqrt{3})+{y^2\over2}\arctan''(\xi)$$
for some $\xi$ in the interval $\bigl[{3\over2},\sqrt{3}\bigr]$. As $\arctan''$ is increasing in this interval we have $$\arctan''(\xi)>\arctan''\bigl({3\over2}\bigr)=-{48\over169}>-{3\over10}\ .$$ It follows that
$$\arctan\left({\pi\over2}\right)>{\pi\over3}-{y\over4}-{3 y^2\over20}>{\pi\over3}-{85\over1922}\doteq1.00297\ .$$
$$ $$
In the first  attempt it was proven that $\tan 1< {\pi\over2}$. I leave it here:
Put
$$x:={\pi\over3}-1\ ,\qquad t:=\tan x\ .$$
Using continued fractions one easily proves that
$${5\over 106}<x<{1\over21}\ .$$
 Then
$$\tan 1=\tan\left({\pi\over3}-x\right)={\sqrt{3}-t\over1+\sqrt{3} t}\ .$$
Now
$${\sqrt{3}-t\over1+\sqrt{3} t}\leq(\sqrt{3}-t)(1-\sqrt{3}t+3t^2)=\sqrt{3}-4t+4\sqrt{3}t^2=:f(t)\ .$$
For  $t\ll 1$ the function $f$ is decreasing. Therefore we can write
$$\tan 1\leq f(t)\leq f(x)\leq f\left({5\over 106}\right)\doteq 1.55879<{\pi\over2}\ .$$
A: I'm not sure if this would be considered nice, but anyway:
It suffices to show that $\tan{1} \leq \frac{\pi}{2}$, or equivalently $\sin{1} \leq \frac{\pi}{2}\cos{1}$.
$$\sin{x}=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!})+\sum_{n=2}^{\infty}{\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right)}$$
$$\cos{x}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!})+\sum_{n=2}^{\infty}{\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)}$$
$$(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!})<\frac{\pi}{2}(1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!})$$
reduces to $3.1+\frac{40.7}{2723}=\frac{8482}{2723}<\pi$, which is true.
$$\left(\frac{1}{(4n+1)!}-\frac{1}{(4n+3)!}\right)<\frac{\pi}{2}\left(\frac{1}{(4n)!}-\frac{1}{(4n+2)!}\right)$$
reduces to $(4n+2)(4n+3)-1<\frac{\pi}{2}((4n+1)(4n+2)(4n+3)-(4n+3))$, which is true.
Thus $\sin{1} \leq \frac{\pi}{2}\cos{1}$, so we are done.
