Proving the inequality $\arctan\frac{\pi}{2}\ge1$

Solution 1:

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 ${\pi \over 2} - 1 = 0.5707963$.

Solution 2:

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: $$\arctan\frac{\pi}{2}\geq 1.$$