How do I show $f(x+2)-f(x)>2 \forall x$? For the function 

$f(x)=x\cos{\frac{1}{x}}$, $x\geq1$, 

How do I show that $f(x+2)-f(x)>2 \forall x$?

$\cos{\frac{1}{x}}=1-\frac{1}{x^22!}+\frac{1}{x^44!}-....$
$x\cos{\frac{1}{x}}=x-\frac{1}{x2!}+\frac{1}{x^34!}-....$
So $(x+2)\cos{\frac{1}{x+2}}=(x+2)-\frac{1}{(x+2)2!}+\frac{1}{(x+2)^34!}-....$
So $f(x+2)-f(x)=2+\{\frac{1}{2x}-\frac{1}{2(x+2)}\}+...$
$=2+\frac{1}{x(x+2)}+...>2$
NOTE: As the number of terms in the summation increases  the magnitude of the terms decreases in particular less than the term before and since $x>0$ the terms are always positive .
