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 $f(x+2)-f(x)>2 \forall x$?
 A: Let $g(x) = x$
Assume $x > 1$
$f(x+2)−f(x)>2$
$=>$$(f(x+2)-g(x+2))$−$(f(x)-g(x))>0$
Now if we can prove $f-g$ is monotonically increasing we are done.
$f-g$ = $x(Cos(1/x)-1)$
Now we want to prove that the derivative is always positive.
Take derivative we get:
$D(f-g)$ = $-1 + Cos(1/x) + Sin(1/x)/x$
Notice that the derivative is $0$ at $Infinity$
Then, if we can show that the derivative is decreasing we can show that the derivative is positive at $x > 1$.
Take derivative again:
D(D(f-g)) = $-(Cos[1/x]/x^3)$, which gives the desired result.
Done
A: First, by the Mean Value Theorem there is an $x_0\in(x,x+2)$ such that $f^\prime(x_0)=\frac{f(x+2)-f(x)}{2}$. But $f^\prime(y)=\frac{1}{y}\sin\frac{1}{y}+\cos\frac{1}{y}$. If we can show that $f^\prime(y)\ge 1$ for all $y\ge 1$, we finish.
Taking $u=1/y$, the last condition is equivalent to show that $u\sin u+\cos u\ge1$ for $u\in(0,1]$. Let $g(u)=u\sin u+\cos u$. Then $g^\prime(u)=u\cos u>0$ for $u\in(0,1]$. Then, $g$ is an increasing function in $(0,1]$. By continuity, this implies that, for $u\in (0,1]$, $g(u)> \lim_{t\to 0^+}g(t)=1$.
Conclusion: $g(u)> 1$ for all $u\in(0,1]$, and this implies $f^\prime(y)< 1$ for all $y\ge 1$. In particular $f^\prime(x_0)< 1$. 
A: $\cos{\frac{1}{x}}=1-\frac{1}{x^22!}+\frac{1}{x^44!}-....$
$x\cos{\frac{1}{x}}=x-\frac{1}{x2!}+\frac{1}{x^34!}-....$
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 .
