Non-negativity of a complicated function

Question

Show that $f(x)\ge 0$ for $0\le x \le 1$, where:

$$f(x) = \arccos(x)^2 -8x(5x^2-2) \sqrt{1-x^2}\arccos(x)+36 x^8-112 x^6+93 x^4-17 x^2$$

The endpoints are $f(0)=\pi^2/4$ and $f(1)=0$. Plotting verifies that the function is positive, but it is not monotonic, or concave. The maximum is the root of a complicated function, so it is not clear how to separate into sub-intervals. There seems to be no way to separate the function into the sum of two simpler parts that are positive.

The problem is equivalent to showing that $\log\hspace{-1pt}\big( h(r)\big)$ is concave on $(0,1]$, where:

$$h(r) = r^{-1}\Big(\pi -2 (1-2r ) \sqrt{r}\sqrt{1-r}-2 \arccos\big(\hspace{-2pt}\sqrt{r}\big)\Big),\quad 0< r \le 1 $$

Update: based on a comment, the problem can be reduced to showing that $g(x) \le 0$ for $1/2 \le x \le 1$, where:

$$g(x) = x\big(x^6-\mbox{$\frac{7}{3}$} x^4 + \mbox{$\frac{103}{72}$}x^2-\mbox{$\frac{25}{144}$}\big)\sqrt{1-x^2}+\mbox{$\frac{5}{9}$}\big(x^4-\mbox{$\frac{19}{20}$}x^2+\mbox{$\frac{7}{80}$}\big)\arccos(x)$$

Since $g$ is monotonic, this is equivalent to showing that $w(x)\ge 0$ on $[1/2,1]$, where:

$$w(x) = x(40 x^2-19)\sqrt{1-x^2}\arccos(x)-144 x^8+378 x^6-323 x^4+93 x^2-4$$

This may seem somewhat obscure, but it turns out to be important for proving a significant theorem on uniqueness of solutions to unmixing linear combinations of independent random variables. One part of the solution I posted in another question. I believe I have a solution for that, though it involves working with a piecewise function and verifying monotonicity of 60 essentially elementary functions. The other part, involving the integral over the radial direction comes down to proving that this function is positive (derived and extracted from a more complicated expression).

Does anyone know how to show that a function like this is positive, or have an idea that I could pursue? I'm asking here because I have no idea how to proceed.

Answer 1

We have to show that \begin{equation*} g\overset{\text{(?)}}\le0 \text{ on }[1/2,1]. \end{equation*} Note that $p(x):=x^4-\frac{19}{20}x^2+\frac{7}{80}$ is of the same sign as $x-x_*$ for $x\in[1/2,1]$, where $x_*:=\frac{1}{2} \sqrt{\frac{1}{10} \left(19+\sqrt{221}\right)}=0.920\dots$.

For $x\in[1/2,1]\setminus\{x_*\}$, let \begin{equation*} h(x):=\frac{g(x)}{p(x)}. \end{equation*} It suffices to show that \begin{equation*} h\overset{\text{(?)}}\ge0 \text{ on }[1/2,x_*) \tag{10}\label{10} \end{equation*} and \begin{equation*} h\overset{\text{(?)}}\le0 \text{ on }(x_*,1]. \tag{20}\label{20} \end{equation*} For $x\in[1/2,1)\setminus\{x_*\}$, \begin{equation*} h''(x)=\frac{x h_2(x)}{28800 p(x)^3\,\sqrt{1-x^2}}, \end{equation*} where \begin{equation*} h_2(x):=-345600 x^{14}+1266240 x^{12}-1823120 x^{10}+1301740 x^8-478566 x^6+85096 x^4-5777 x^2-497<0 \end{equation*} for $x\in[1/2,1]$.

So, $h$ is convex on $[1/2,x_*)$ and concave on $(x_*,1]$.

We have $h(1)=0=h'(1)$. So, \eqref{20} follows.

Also, $h'(3/4)=0.119\ldots>0$ and hence for all $x\in[1/2,x_*)$ we have $h(x)\ge h(3/4)+h'(3/4)(x-3/4) \ge h(3/4)+h'(3/4)(1/2-3/4)=0.102\ldots>0$. So, \eqref{10} follows as well. $\quad\Box$

Answer 2

If you can bound $f'(x)$, you only need to numerically evaluate finitely many points to show that the function is positive everywhere. (Given $-A < f'(x) < A$, if $f(x_0) = C$, take $x_1 = x_0 + C/A$.)

This doesn't work at $f(1) = 0$, but showing the negativity of $f'(x)$ around that point (between the last x-value you check and x=1) would suffice. Unfortunately, $f'(1) = 0$, so you'd need to show the positivity of $f''(x)$ near $1$. $f''(x)$ is positive at $x=1$, so the original strategy will work for it. It just unfortunately means you'd have to bound $ f'''(x)$.

Answer 3

Proof.

We split into two cases.

Case 1: $5x^2 - 2 \le 0$

Using $\arccos x \ge \sqrt{1 - x^2}$ for all $x\in [0, 1]$ (equivalently $y \ge \sin y$ by simply letting $x = \cos y$), we have \begin{align*} &f(x)\\ \ge{}& (1 - x^2) - 8x(5x^2 - 2)\sqrt{1 - x^2} \cdot \sqrt{1 - x^2} + 36x^8 - 112 x^6 + 93x^4 - 17x^2\\ ={}& \left( x+1 \right) \left( 36\,{x}^{5}+36\,{x}^{4}-40\,{x}^{3}+17\,x+ 1 \right) \left( x-1 \right) ^{2}\\ \ge{}& 0. \end{align*}

Case 2: $5x^2 - 2 > 0$

Using $\arccos x \ge \sqrt{1 - x^2}$ for all $x \in [0, 1]$ and $\arccos x \le \frac{\sqrt{1 - x^2}}{x}$ for all $x \in [0, 1]$ (equivalently $y \le \tan y$ by simply letting $x = \cos y$), we have \begin{align*} &f(x)\\ \ge{}& (1 - x^2) - 8x(5x^2 - 2)\sqrt{1 - x^2} \cdot \frac{\sqrt{1 - x^2}}{x} + 36x^8 - 112 x^6 + 93x^4 - 17x^2\\ ={}& \left( 36\,{x}^{4}-40\,{x}^{2}+17 \right) \left( x-1 \right) ^{2} \left( x+1 \right) ^{2} \\ \ge{}& 0. \end{align*}

We are done.