For $x\geq 2$,prove that$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$ For $x\geq 2$,prove that$$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$$
My Attempt:
Let $f(x)=x\cos\left(\frac{\pi}{x}\right)$
$$f'(x)=\cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)>0$$for all $x\in[2,\infty)$
$f''(x)=-\frac{{\pi}^2}{x^3}\cos\left(\frac{\pi}{x}\right)<0$ for all $x\in[2,\infty)$
By LMVT on $[x,x+1]$,we have
$$f(x+1)-f(x)=\cos\left(\frac{\pi}{c_{x}}\right)+\frac{\pi}{c_{x}}\sin\left(\frac{\pi}{c_{x}}\right)$$for some $c_{x}\in(x,x+1)$
After this I am not able to go ahead.
 A: According to your work, 
$$f''(x)=-\frac{{\pi}^2}{x^3}\cos\left(\frac{\pi}{x}\right)<0$$ for all $x\in(2,+\infty)$. Hence
$$f'(x)=\cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)$$ 
is strictly decreasing in $(2,+\infty)$. Since $\lim_{x\to +\infty}f'(x)=1$, it follows that $f'(t)>1$ for all $t> 2$. Finally, for $x\geq 2$, by the Mean Value Theorem, there is $t\in (x,x+1)\subset (2,+\infty)$ such that
$$f(x+1)-f(x)=f'(t)(x+1-x)=f'(t)>1.$$
A: You're done if you prove that $f'(x)>1$ for $x>2$. By setting $\pi/x=t$, this is the same as proving that
$$
g(t)=\cos t+t\sin t>1,\qquad 0<t<\pi/2
$$
Note that $g'(t)=t\cos t>0$ and $\lim_{t\to0}g(t)=1$.
A: Another possible way of proving your inequality is rearranging and then applying the MVT as follows:
\begin{eqnarray*}
(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right) & > & 1\\
& \Leftrightarrow & \\
x\left( \cos \frac{\pi}{x+1} - \cos \frac{\pi}{x} \right) & > & 1 - \cos \frac{\pi}{x+1} \\
& \stackrel{MVT:\, \color{blue}{\frac{\pi}{x+1}<\xi < \frac{\pi}{x},\, 0< \zeta < \frac{\pi}{x+1}}}{\Longleftrightarrow} & \\
x\sin \xi \cdot \left(\frac{\pi}{x} - \frac{\pi}{x+1} \right) = \color{blue}{\sin \xi} \cdot \frac{\pi}{x+1} & > & \color{blue}{\sin \zeta} \cdot \frac{\pi}{x+1} \\
\end{eqnarray*}
Now note that $\sin x$ is strictly increasing on $[0,\frac{\pi}{2}]$. Hence, for $x\geq 2 \Rightarrow \color{blue}{\frac{\pi}{2}> \xi > \zeta > 0}\Rightarrow \color{blue}{\sin \xi > \sin \zeta}$. So, the last inequality is $\color{blue}{\mbox{true}}$ for $x\geq 2$.
