Show that $f(x)=x+\frac{1}{x^2+1}$ is strictly increasing. Show that $f(x)=x+\frac{1}{x^2+1}$ is always increasing.
I noticed that $f'(x)=1-\frac{2x}{(x^2+1)^2}$, by looking at this function graph, I noticed it must be always positive, however I can't find a way to prove that. My attempt:
If $x\le0$, then clearly $f'(x)>0$. If $x\ge1$, it is also quite evident that $\frac{2x}{(x^2+1)^2}\le1$. Then, when $0<x<1$, I'm a bit lost. I can't seem to find an algebraic trick to show that $2x<(x^2+1)^2$, I'd love some help.
 A: To complete your argument:
$$
(x^2+1)^2=x^4+2x^2+1=(x^4+x^2)+(x^2+1).
$$
Now note that $(x^4+x^2)\geq 0$ and $x^2+1\geq 2x$ but equalities cannot happen simultaneously. This argument actually shows $2x<(x^2+1)^2$ always, so you don't need to split into cases.
You can also do a calculus-free proof:
$$
f(y)-f(x)=(y-x)\frac{(x-0.5)^2+(y-0.5)^2+0.5+x^2y^2}{(1+x^2)(1+y^2)}.
$$
A: In fact, if $\frac 12\leq x <1$, then $(2x) \leq (2x)^2 \leq (x^2+1)^2$. While if $0< x < \frac 12$, we clearly have $2x \leq 1 \leq (x^2+1)^2$. This is the simplest proof that I can come up with
A: There is a common result stating that $\quad 2ab\le a^2+b^2\quad$  (just develop $(a-b)^2\ge 0$).
In particular for $a=1$ and $b=x$ you get $\, 2x\le 1+x^2$
And since $1+x^2\ge 1$ then $1+x^2\le (1+x^2)^2$ and your result is proved.
A: $\theta = \arctan (x), g(\theta): = f(x) = \tan(\theta)+ \cos^2 (\theta)$
$g'(\theta) = \sec^2 \theta - \sin (2\theta) \ge 1-1=0$. (Can't have equality because if $\sec^2\theta=1$ then $\cos\theta=\pm1, \sin2\theta=0$.)
And $x$ increases with $\theta$.
