Monotonicity of some functions. Let $f:[0,\infty)\to[0,\infty)$ be a decreasing function, decreases to $0$ as $x \to ∞$. 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: The answer is no. 
To construct $f$, start by writing 
$$
[0,\infty)=\bigcup_{n=0}^\infty [n+\delta_n,n+1+\delta_{n+1})
$$
Here, $\delta_0=0$ and, for $n\ge 1$, each $\delta_n\in (0,1)$ is small enough so that the slope of the line joining the points $\left(n,\frac{1}{n}\right)$ and $\left(n+\delta_n,\frac{1}{n+1}\right)$ is smaller than $-n$. In other words,
$$
0<\delta_n<\frac{\frac{1}{n}-\frac{1}{n+1}}{n}
$$
Now define $f$ as follows:


*

*On $[0,1)$, $f=1$. 

*On $[1,\delta_1)$, the graph of $f$ is the line segment joining the points $(1,1)$ and $\left(1+\delta_1,\frac{1}{2}\right)$. 

*On $[1+\delta_1,2)$, $f=\frac{1}{2}$.

*On $[2,2+\delta_2)$, the graph of $f$ is the line segment joining the points $\left(2,\frac{1}{2}\right)$ and $\left(2+\delta_2,\frac{1}{3}\right)$.


etc
It follows that, for $x\in (n,n+\delta_n)$, we have that $xf'(x)<-1$. 
Finally, if $g(x)=xf(x)+\sqrt x$, then
$$
g'(x)=f(x)+xf'(x)+\frac{1}{2\sqrt x}
$$
(except on a countable set, but that's irrelevant). Therefore, $g'(x)<0$ for $x\in (n,n+\delta_n)$ for all large enough $n$.
