Show monotonicity Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \  \ \text{for} \ \  x, a,b,c > 0 \ .
$$ 
To do so, I figured it is sufficient to show that
$$
f(x) = \sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)
$$
is monotonically increasing for $x > 0$. 
Of course, I took the derivative $f'(x)$ and proceeded with the demand 
$$
\frac{\frac{b}{a}x}{\sqrt{a \cdot b + \frac{b}{a}x^2}} \arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) - \frac{c \frac{b}{a} x}{a \cdot b + \frac{b}{a} x^2 + c^2} > 0 \ .
$$
In the end, I got stuck with
$$
\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) > \frac{c \sqrt{a \cdot b + \frac{b}{a}x^2}}{a \cdot b + \frac{b}{a} x^2 + c^2} \ .
$$
Inserting values for a,b, and c seems to work perfectly, but I can't manage to analytically solve the inequation.
Does anyone have an idea of how to approach this problem? 
Edit:
I came across the Shafer-Fink inequality stating
$$
\frac{3y}{1+2\sqrt{1 + y^2}} < \arctan y < \frac{\pi y}{1 + 2\sqrt{1 + y^2}} \ .
$$
Can I substitute 
$$
y = \frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}
$$
and therefore receive 
$$
\arctan (y) > \frac{3y}{1+2\sqrt{1 + y^2}} > \frac{y}{1 + y^2} ?
$$
Is that a proper way?
 A: There is an integral for arctan(x):
\[
\arctan(x) = \int_{0}^{x} \frac{dt}{t^2 + 1}, \quad x > 0.
\]
Perhaps this can help.
A: I'll first rearrange and reparametrise the original equality a little to make it easier to work with. Let $\alpha := c/\sqrt{ab}$, $z:= \sqrt{ 1 + (x/a)^2}$. Since $a,b,c >0$, it should be easy to see that the original inequality is equivalent to $$ z \arctan \frac{\alpha}{z} \overset{?}\ge \arctan \alpha$$ for every $\alpha >0, z> 1$ - much neater.
Note that the relation above is true for $z=1$. Now, going by your proof technique, we'll be done if we manage to show that $$ \arctan \frac{\alpha}{z} \ge \frac{\alpha/z}{1 + (\alpha/z)^2}$$ for every $\alpha >0, z>1$. But this is true since, for $\alpha/z>0$, $$\arctan \frac{\alpha}{z} = \int_0^{\alpha/z} \frac{1}{1+t^2} \mathrm{d}t \overset{(1)}\ge \int_0^{\alpha/z}\frac{1}{1+(\alpha/z)^2} \mathrm{d} t = \frac{\alpha/z}{1+(\alpha/z)^2},$$ where the inequality $(1)$ holds since $1+t^2$ is an increasing function for $t>0$.
