Prove that a function have no maximum

Question

I want to prove that $\frac{\arctan(x)}{x} $ have no maximum in the interval $(0,\infty)$ ?

I did proved that $\lim \limits_{ x \to 0+} \frac{\arctan(x)}{x} =1$ and $\lim \limits_{x \to \infty}\frac{\arctan(x)}{x}=0$

And did assume by contradiction that $ \frac{\arctan(x)}{x}$ is the maximum value and i want to prove that $\frac{\arctan(x/2)}{x/2}$ is bigger, but here were i get stuck !

Answer 1

Well, with $\theta=\arctan(x/2)$, you have $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2(x/2)}{1-(x/2)^2}>x$ for $0<x/2<1$, i.e. $\arctan x<2\arctan(x/2)$. For $x/2\ge1$, the inequality is valid because $\arctan x<\pi/2\le2\arctan(x/2)$.
Alternatively, you could use the integral representation $$\frac{\arctan x}x=\frac1x\int^x_0\frac1{1+t^2}\,dt=\int^1_0\frac1{1+x^2u^2}\,du$$ to see that your function is monotone decreasing for $x>0$.