# Find the best constant in a function inequality.

Find the best constants $c_1$ and $c_2$ such that: $$\frac{c_1x}{1+x^2}\leq \arctan(x)\leq \frac{c_2x}{1+x}\qquad \forall x\geq0$$

What I've tried so far: It's easy to determine the constant $c_1$ just by studying $f(x)=\frac{(1+x^2)\arctan(x)}{x}$.

By analyzing this function we can see that it has a lower bound which is $1$, so the best constant is clearly $c_1=1$. But what about the second inequality? I tried to do the same by studying $g(x)=\frac{(1+x)\arctan(x)}{x}$ but I really have difficulties in finding the maximum of the function (which I know, by computation, exists).

Any hint/help is really appreciated.

• NoTe : taht ....left hand side is wrong for $-1<x<0$ !!! Aug 22 '17 at 9:54
• Oh, sorry, I forgot to say for all $x\geq 0$. I'll edit. Aug 22 '17 at 9:56

For any $c\in(1,2)$, the function $f_c(x)=\frac{c x}{1+x}-\arctan(x)$ attains its maximum value at $x=\frac{1+\sqrt{2c-c^2}}{c-1}$. Let $d=\frac{1+\sqrt{2c-c^2}}{c-1}$: we have $d\in(1,+\infty)$ and $c=\frac{(1+d)^2}{1+d^2}$, so the problem boils down to finding a zero of $$g(d)=\frac{(1+d)^2}{1+d^2}\cdot\frac{d}{1+d}-\arctan(d)=\frac{d(1+d)}{1+d^2}-\arctan(d)$$ on $(1,+\infty)$, or, by setting $d=\tan\theta$, a zero of $$h(\theta) = \sin\theta\,(\sin\theta+\cos\theta)-\theta$$ on the interval $\theta\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$. If we apply a step of Newton's method to $\frac{h(\theta)}{\theta}$ with starting point $\theta_0=\frac{\pi}{2}$, we get that the only zero of $h(\theta)$ is close to $\frac{2\pi}{2+\pi}$, hence:
$$c_2 \approx \color{blue}{1+\sin\left(\tfrac{4\pi}{2+\pi}\right)}\approx\tfrac{5}{3}.$$
The above substitutions show that the problem of finding $c_2$ is essentially equivalent to solving Kepler's equation, and there is a large amount of literature on such topic. See, for instance, Colwell - Solving Kepler's Equations Over three Centuries.
You are not going to find the exact value of $c_2$ I think. You should study the function $f_a(x)=\frac{ax}{1+x}-\arctan x$. Since $$\lim_{x\to\infty}\frac{ax}{1+x}-\arctan x=a-\frac\pi2$$ you want $a\ge\frac\pi2$. Computing the derivative of $f_a$ you get that $$\frac{d}{dx}\left( \frac{ax}{1+x}-\arctan x\right) =\frac{1}{\left( x^{2}+1\right) \left( x+1\right) ^{2}}\left( (a-1)x^{2}-2x+a-1\right)\ge 0$$ for $(a-1)x^{2}-2x+a-1\geq0$. This equation has two distinct solutions for $0<a<2$, which are $$\frac{1\pm\sqrt{1-(a-1)^{2}}}{a-1}.$$ Hence, the function increases for $0<x<\frac{1-\sqrt{1-(a-1)^{2}}}{a-1}$ and decreases for $x>\frac{1+\sqrt{1-(a-1)^{2}}}{a-1}$. This shows that at $\frac{1+\sqrt{1-(a-1)^{2}}}{a-1}$ you have a global minimum. You want the function $f_a$ to be always positive, so you are interested in the value of $f_a$ at $\frac{1+\sqrt{1-(a-1)^{2}}}{a-1}$. Define $$g(a):=f_a\left(\frac{1+\sqrt{1-(a-1)^{2}}}{a-1}\right)=\frac{a+a\sqrt{1-(a-1)^{2}}}{a+\sqrt{1-(a-1)^{2}}}-\arctan\frac {1+\sqrt{1-(a-1)^{2}}}{a-1}%$$ For $a=\frac\pi2$ you have $g(a)=1.1959- 1. 5708<0$ while for $a=2$ you have $g(2)=1-\arctan 1>0$. Since $g$ is continuous, there exists a first value of $a$ after $\frac\pi2$ such that $g(a)=0$. This gives you the best constant.
We certainly need $c:=c_2\geq{\pi\over2}$. The function $$f(x):={cx\over 1+x}-\arctan x\qquad(x\geq0)$$ has derivative $$f'(x)={(c-1)x^2-2x+(c-1)\over(1+x)^2(1+x^2)}\ ,$$ hence increases for small $x\geq0$ to a local maximum, then decreases to a local minimum, and from then on increases indefinitely, see the following figure. The minimum is at $x_*=\bigl(1+\sqrt{2c-c^2}\bigr))/(c-1)$. We now have to find the smallest $c$ for which $f(x_*)\geq0$. Due to the $\arctan$ this leads to a transcendental equation, so that we have to settle with a practical numerical value. It so happens that $c={5\over3}$ leads to $x_*=(3+\sqrt{5})/2$, and above all to $f(x_*)\approx 0.0000788309>0$.