Finding lower bounds for a function Given a function $w\left(x\right)=\sum_{n=1}^x\lfloor\frac{x}{n}\rfloor$.
And $f(x) = w(6x+1)-w(6x-2)$, prove that for $x>3; f(x) \geq 12$
EDIT: This doesn't really come from anywhere, I was just messing around with discrete functions and wondered if I could show that there are an infinite amount of instances where f(x) is equal to 12. What I tried was manually counting the trivial values. For example for n = 1 we would already have 3 and by nature $w(6x+1)$ will be 3 more than $w(6x-2)$. 
 A: You have the $2$ functions
$$w(x) = \sum_{n=1}^x \left\lfloor \frac{x}{n} \right\rfloor \tag{1}\label{eq1}$$
\begin{align}
f(x) & = w(6x + 1) - w(6x - 2) \\
& = \sum_{n=1}^{6x+1} \left\lfloor \frac{6x+1}{n} \right\rfloor - \sum_{n=1}^{6x-2} \left\lfloor \frac{6x-2}{n} \right\rfloor \\
& = \left\lfloor \frac{6x+1}{6x+1} \right\rfloor + \left\lfloor \frac{6x+1}{6x} \right\rfloor + \left\lfloor \frac{6x+1}{6x-1} \right\rfloor + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \\
& = 3 + \sum_{n=1}^{6x-2} \left(\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor\right) \tag{2}\label{eq2}
\end{align}
Consider any integer $x \gt 3$. In the summation part of \eqref{eq2}, each term $\left\lfloor \frac{6x+1}{n} \right\rfloor - \left\lfloor \frac{6x-2}{n} \right\rfloor \ge 0$. Also, among these terms, for $n = 1$, you get $(6x+2) - (6x-1) = 3$; for $n = 2$, you get $(3x) - (3x-1) = 1$; for $n = 3$, you get $(2x) - (2x-1) = 1$; for $n = 6$, you get $x - (x-1) = 1$; for $n = x$, you get $6 - 5 = 1$; for $n = 2x$, you get $3 - 2 = 1$; and for $n = 3x$, you get $2 - 1 = 1$. Note since $x \gt 3$, each of these values of $n$ being used (i.e., $1,2,3,6,x,2x,3x$) are distinct, except for where $x = 6$. In this case, you can replace the $n$ value of $x$ with $n = 4$ instead since it has $9 - 8 = 1$.
Adding these parts together gives $3 + 1 + 1 + 1 + 1 + 1 + 1 = 9$. Thus, along with the initial $3$ in \eqref{eq2} and that each term in the sum is non-negative, this shows the overall value must be at least $12$, i.e., $f(x) \ge 12$, for all $x \gt 3$.
