An application of derivatives problem- to show that $f(x+2)-f(x)>2$ for $f(x)=x\cos(1/x)$

Question

I have this function $$f(x) = x\cos(1/x)$$ defined for all $x \geq 1$.

How do I prove that $$f(x+2)-f(x)>2$$ for all $x\geq 1$?

I found $$f'(x)= \cos(1/x) + \frac{\sin(1/x)}{x}$$ and $$f''(x)=-\frac{\cos(1/x)}{x^3}.$$ How do i use this information?

Answer 1

$f''(x) <0$ so $f'(x)$ is decreasing. min value of $f'(x)$ occurs at $\infty$, which is $1$. now apply LMVT on $[x,x+2]$, $f'(c) = \frac{f(x+2)-f(x)}{x+2-x}$ but $f'(c) > 1$, for $x> 1$ hence proved