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.
 A: Potato's answer is what's going on geometrically. If you want it analytically:$$\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2} \geq \sqrt{\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}$$
$$ = \bigg|\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\bigg|$$
The above expression is the absolute value of the derivative of $t\sin(\pi/t)$. 
So your integral is greater than 
$$\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}|(t\sin({\pi \over t}))'|\,dt + \int_{1 \over n + {1 \over 2}}^{1 \over n}|(t\sin({\pi \over t}))'|\,dt$$
This is at least what you get when you put the absolute values on the outside, or
$$\bigg|\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}(t\sin({\pi \over t}))'\,dt\bigg| + \bigg|\int_{1 \over n + {1 \over 2}}^{1 \over n}(t\sin({\pi \over t}))'\,dt\bigg|$$
Then the fundamental theorem of calculus says this is equal to the following, for $f(t) = t \sin(\pi/t)$:
$$\bigg|f({1 \over n + {1 \over 2}}) - f(0)\bigg| + \bigg|f({1 \over n}) - f({1 \over n + {1 \over 2}})\bigg|$$
$$= \bigg|{1 \over n + {1 \over 2}} - 0\bigg| + \bigg|0 -{1 \over n + {1 \over 2}}\bigg|$$
$$ = {2 \over n + {1 \over 2}}$$
A: 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).
A: Now that I have finished this, I see that it is similar to the approach that Zarrax used, but it looks a bit simpler, so I will post it in addition.
Using the following facts
$$
\sqrt{x^2+1}\ge|x|\tag{1}
$$
$$
|x-y|\ge\mathrm{sgn}(x)(x-y)\tag{2}
$$
with a change of variables $t\mapsto 1/t$, we get
$$
\begin{align}
&\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}\mathrm{d}t\\
&=\int_n^{n+1}\sqrt{1+(\sin(\pi t)-\pi t\cos(\pi t))^2}\frac{\mathrm{d}t}{t^2}\\
&\ge\int_0^1\left|\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right|\mathrm{d}t\\
&\ge\int_0^1\mathrm{sgn}(\cos(\pi t))\left(\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right)\mathrm{d}t\\
&=\int_0^1\mathrm{sgn}(\cos(\pi t))\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\sin(\pi t)}{n+t}\right)\mathrm{d}t\\
&=\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(0)}{n}\right)+\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(\pi)}{n+1}\right)\\
&=\frac{2}{n+\frac12}\tag{3}
\end{align}
$$
A: Hint: notice that the integral may be written as: 
$$\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt= \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+{\left[\left(t\sin\left(\frac{\pi}{t}\right)\right)'\right]}^2}dt=\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1 + [f(t)']^2}dt$$ where $f(t)= t \sin\frac{\pi}{t}.$ That means that you need to prove that the lenght of the graph of the function $f(t)$ from $\frac{1}{n+1}$ to $\frac{1}{n}$ is greater than or equal to $\frac{2}{n+\frac{1}{2}}$. What does it happen with the function 
$f(t)= t \sin\frac{\pi}{t}$ at the points $\frac{1}{n+1}$ and $\frac{1}{n}$?
Geometrically, the result is evident.
