Trying to prove $\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+(\sin(\frac{\pi}{t}) -\frac{\pi}{t}\cos(\frac{\pi}{t}))^2}dt$ I posted this incorrectly several hours ago and now I'm back! So this time it's correct.  Im trying to show that for $n\geq 1$:
$$\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt$$
I checked this numerically for several values of $n$ up through $n=500$ and the bounds are extremely tight.  
I've been banging my head against this integral for a while now and I really can see no way to simplify it as is or to shave off a tiny amount to make it more palatable.  Hopefully someone can help me.  Thanks.

It is not hard to show (for example, in the problem following the problem you posted, exercise 10 in Section 1-3 of Differential Geometry of Curves and Surfaces by Do Carmo) that the arc length of and arc with endpoints $x$ and $y$ is at least the length of the straight line segment connecting them. In any case, the problem only asks for a "geometrical" proof. 
That integral is the arc length of the curve $f(t) =(t,\sin (\pi/t))$ between the points $t=1/(n+1)$ and $t=1/n$. These points are $(1/(n+1), \sin((n+1)\pi)/(n+1))$ and $(1/n, \sin(n\pi)/n)$ (so the $y$ coordinates are $0$). Call them $A$ and $B$, respectively. The arc passes through the point $(1/(n+1/2),\sin(n\pi/2)/(n+(1/2))=(1/(n+1/2),\pm 1/(n+(1/2))$. Call this $C$. We see the arc length is at least the sum of the length of the segments $AC$ and $CB$. These each have length at least $\frac{1}{n+\frac{1}{2}}$ (draw the picture. This is the length of the perpendicular to the $x$-axis).
