I Is the Arctan Convergence Rate Claim Valid for Positive Values of a, b, and x?

[jostpuur]

What do you think about the claim that

<br /> \frac{x}{\frac{1}{a} + \frac{x}{b}} \;&lt;\; \frac{2b}{\pi}\arctan\Big(\frac{\pi a}{2b}x\Big),\quad\quad\forall\; a,b,x&gt;0<br />

First I thought that if this is incorrect, then it would be a simple thing to find a numerical point that proves it, and also that if this is correct, then it would be a simple thing to prove this via some derivatives or integrals. For some reason I didn't succeed in either objective, and now I'm not sure why I feel confused about this.

 

[I like Serena]

Hi jostpuur,

Since we can scale the graphs horizontally and vertically, without loss of generality, we can assume that a=1 and b=1.
Let:
$$f(x)=\frac 2\pi \arctan\left(\frac \pi 2 x\right) - \frac x{1+x}$$
Then:
$$f'(x)=\frac{1}{1+(\frac\pi 2 x)^2} - \frac 1{(1+x)^2} = \frac{x(8-(\pi^2-4)x)}{(4+\pi^2 x^2)(1+x)^2}$$
The derivative is zero at $x=0$ and $x=\frac{8}{\pi^2-4}$.
And we have a horizontal asymptote at $y=0$ when $x\to\infty$.
It follows that the graph $y=f(x)$ has a minimum at $x=0$, a maximum at $x=\frac{8}{\pi^2-4}$, and then descends to its asymptote $y=0$.
QED

 

[jostpuur]

I see, thanks. If f reached negative values, then f&#039; should have a third zero somewhere.

You made a typo with the quantity \big(\frac{\pi}{2}x\big)^2 in an intermediate step.