About monotonicity of a function. Let $f:[0,\infty)\to[0,\infty)$ be a decreasing function, decreases to $0$ as $x \to \infty$ and
$$\int_1^\infty \dfrac{f(x)}{x}=\infty.$$
Can we find $M\geq 0$ such that 
$$g(x)=xf(x)+\sqrt{x}$$
is increasing on $[M,\infty)$ and $g(x)\to \infty$ as $x\to \infty$?
 A: Consider a sequence $x_1,x_2,\ldots $ strictly increasing from $x_1=1$ to $\infty$ and a sequence $y_1, y_2,\ldots$ strictly decreasing to $0$.
Then defining 
 $$f(x)=y_n\qquad \text{for }x_n\le x<x_{n+1}$$
gives us a weakly decreasing function $[1,\infty)$ to $[0,\infty)$ with $f(x)\to 0$ as $x\to\infty$.
Moreover, 
$$
\int_1^\infty \frac{f(x)}x\,\mathrm dx= \sum_{n=0}^\infty \int_{x_n}^{x_{n+1}} \frac{f(x)}x\,\mathrm dx\ge \sum_{n=0}^\infty \left(1-\frac {x_n}{x_{n+1}}\right)y_n.$$
Hence to satisfy the second condition, we might for example pick $x_n=2^{n-1}$ and $y_n=\frac2n$ to make  all summands $\ge \frac1{n}$ and hence obtain the divergent harmonic series.
Now for such $f$, the function $g$ will have downward jump discontinuities at every $x_n$, in particular infinitely many in $[M,\infty)$, no matter what $M$ we pick.
Admittedly, $f$ as above is only weakly decreasing - but it is easy to correct that by making it only decrease by a very little amount on each subinterval. Also, while the problem statement did not require $f$ to be continuous, even that can be accounted for by repairing the jump discontinuities to short intervals of steep decrease.
