Let $\operatorname{Si}(x)=\int^{x}_{0}\frac{\sin(t)}{t}dt$. Show that $\operatorname{Si}(\pi)>\frac{\pi}{2}$, using the hints $\pi>3$ and $\pi^{2}<10$.
My ideas:
Writing $\frac{sin(t)}{t}$ as a Taylor series (about $0$) we have:
$$T_{n,0}(x)=\sum^{n}_{k=0}(-1)^{k}\frac{\pi^{2k}}{(2k+1)!}$$
Then integrating the Taylor series (about $0$) of $\frac{\sin(t)}{t}$ we find:
$$T_{n,0}(x)=\sum^{n}_{k=0}(-1)^{k}\frac{\pi^{2k+1}}{(2k+1)(2k+1)!}$$
And evaluated at $\pi$ and $0$, in order to get $\operatorname{Si}(x)$, we find:
$$\sum^{n}_{k=0}(-1)^{k}\frac{x^{2k+1}}{(2k+1)(2k+1)!}\bigg\rvert^{\pi}_{0}=\sum^{n}_{k=0}(-1)^{k}\frac{\pi^{2k+1}}{(2k+1)(2k+1)!}$$
In order to use the hints I expanded the series:
$$\sum^{n}_{k=0}(-1)^{k}\frac{\pi^{2k+1}}{(2k+1)(2k+1)!}= \pi-\frac{\pi^{3}}{3\times3!}+\frac{\pi^{5}}{5\times5!}-\frac{\pi^{7}}{7\times7!}+\sum^{n}_{k=4}(-1)^{k}\frac{\pi^{2k+1}}{(2k+1)(2k+1)!}$$
and $\frac{\pi^{2k+1}}{(2k+1)(2k+1)!}$ is monotone decreasing on $k>3$ so all I need to show is $$\pi-\frac{\pi^{3}}{3\times3!}+\frac{\pi^{5}}{5\times5!}-\frac{\pi^{7}}{7\times7!}>\frac{\pi}{2}$$
And using the hints above:
$$\pi-\frac{\pi^{3}}{3\times3!}+\frac{\pi^{5}}{5\times5!}-\frac{\pi^{7}}{7\times7!} > 3-\frac{10\pi}{3\times3!}+\frac{3^{5}}{5\times5!}-\frac{10^{3}\pi}{7\times7!}>4-\frac{\pi}{2}>\sqrt{10}-\frac{\pi}{2}>\sqrt{\pi^{2}}-\frac{\pi}{2}=\pi-\frac{\pi}{2}=\frac{\pi}{2}$$
I believe that my proof although correct is not elegant enough, particularly when using $\sqrt{10}<4$
Any better alternatives, while still using the same hints?
Note that $\frac{\sin(x)}{x}\ge (1-x/\pi)$ for $x\in [0,\pi]$. Therefore, we can write
$$\begin{align} \int_0^\pi \frac{\sin(x)}{x}\,dx&\ge \int_0^\pi (1-x/\pi)\,dx\\\\ &=\pi-\frac12\pi\\\\ &=\frac\pi2 \end{align}$$
as was to be shown!
NOTE:
To show that $\displaystyle \frac{\sin(x)}{x}\ge (1-x/\pi)$ for $\displaystyle x\in [0,\pi]$, it is sufficient to show that the function $\displaystyle f(x)$, given by $\displaystyle f(x)=\sin(x)-x+\frac1\pi x^2$ is non-negative for $\displaystyle x\in [0,\pi/2]$.
We have the following:
$$\begin{align} f'(x)&=\cos(x)-1+\frac2\pi x\\\\ f''(x)&=-\sin(x)+\frac2\pi\\\\ f'''(x)&=-\cos(x) \end{align}$$
For $\displaystyle x\in[0,\pi/2]$, $\displaystyle f'''(x)\le 0$ (i.e., $\displaystyle f'(x)$ is concave).
Inasmuch as $\displaystyle f'(0)=f'(\pi/2)=0$, then $\displaystyle f'(x)\ge 0$ on $\displaystyle [0,\pi/2]$.
And since $\displaystyle f(0)=0$ and $\displaystyle f'(x)\ge 0$ on $\displaystyle [0,\pi/2]$, we see that $\displaystyle f(x)\ge 0$ on $\displaystyle [0,\pi/2]$.