TITLE: Proving the inequality $\arctan\frac{\pi}{2}\ge1$
QUESTION [12 upvotes]: Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.

REPLY [4 votes]: 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 ${\pi \over 2} - 1 = 0.5707963$.