$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$ Let $$f(x) = x\cos\frac{1}{x}$$ for x in $[1,\infty)$
Now I need to prove or disprove the difference $$f(x+2) - f(x) > 2$$ for all $x$ in the domain.
I tried a lot but I don't seem to be getting anywhere. My approach was to try and use graphs, but somehow that doesn't seem to work out that well.
Any ideas or suggestions please?
 A: Note that cosine function is strictly concave on $[0,\frac{\pi}{2}]\supset[0,1]$. It implies that when $x\ge 1$,
$$\frac{2}{x+2}\cos 0 +\frac{x}{x+2}\cos\frac{1}{x} < \cos\left(\frac{2}{x+2}\cdot 0+\frac{x}{x+2}\cdot \frac{1}{x}\right)=\cos\frac{1}{x+2}.$$
The conclusion follows. 
A: Hint: When $x\ge 1$, then $0 \lt \frac 1x \le 1$.  How does $\cos y$ behave when $0\lt y\le 1$?  
A: For an alternative approach consider $g(y) = \cos(y)+y\sin(y)$.
$$g^\prime(y) = -\sin(y)+\sin(y)+y\cos(y) = y\cos(y) >0$$ for $y\in \left(0,\frac{\pi}{2}\right]$. Thus
$$g(y) > g(0) = \cos(0)+0\sin(0) = 1$$ for $y\in \left(0,\frac{\pi}{2}\right]$.
Now calculate $f^\prime(x) = \cos\left(\frac{1}{x}\right)+\frac{1}{x}\sin\left(\frac{1}{x}\right)$. By the previous paragraph, $f^\prime(x)>1$ for $\frac{1}{x}\in \left(0,\frac{\pi}{2}\right]$ or equivalently $x\geq\frac{2}{\pi}$.
Hence, we conclude $\displaystyle f(x+2)-f(x) = \int_x^{x+2} f^\prime(y)\,\mathrm{d}y > \int_x^{x+2} 1\,\mathrm{d}y = 2$ for all $x\geq\frac{2}{\pi}$.
