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Prove the following inequality: $$\log\left(\frac{1+x}{1-x}\right)> \frac{2x\left(x(2x-1) + (3x+4)\log(1+x)\right)}{x(2 + x - x^2) + (4+x-3x^2)\log(1+x)}, \ \mbox{for} \ x\in \left(\frac{1}{2},1\right).$$

Attempt: I used a straightforward approach, but it gets messy. Let $f_1(x)$ denote the LHS, and $f_2(x)$ the RHS. Then, it is easy to show that $f_1(x) =2$ and $f_2(x) = 0$ for $x\to 0$, and $f_1(x) = \infty$ and $(0<)f_2(x) <\infty $ for $x \to 1$. Next, I show that $F(x) \equiv f_1'(x) - f_2'(x)>0$ for both limits. However, it can be $F(x)<0$ for intermediate value of $x$. I tried to show that at the the inequality holds even for negative values of $F(x)$, and this is where I got stuck.

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    $\begingroup$ What is $q$ in the numerator? $\endgroup$ – Shubham Johri Jan 28 '19 at 16:10
  • $\begingroup$ @ShubhamJohri it is supposed to be $x$. Corrected. $\endgroup$ – Bill.Math Jan 28 '19 at 16:12
  • $\begingroup$ The Taylor approximation helps, but we need to work with polynomial of twenty degree. It's true for all $0<x<1$. $\endgroup$ – Michael Rozenberg Jan 28 '19 at 17:57
  • $\begingroup$ How do you make out that $f_2(x)\to\infty$ as $x\to1$? $\endgroup$ – Calum Gilhooley Jan 28 '19 at 19:24
  • $\begingroup$ @CalumGilhooley Thanks for pointing out. Due to you, I also found that I made a mistake in the derivation of the inequality. Still my question remains even after correction. $\endgroup$ – Bill.Math Jan 28 '19 at 20:17
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Write the inequality to be proved in the form $f(x) > g(x)$, where \begin{gather*} f(x) = \frac{1+x}{2x}\log\left(\frac{1+x}{1-x}\right) = (1+x)\left(1 + \frac{x^2}{3} + \frac{x^4}{5} + \cdots\right), \\ g(x) = \frac{(2x-1) + (4+3x)\frac{\log(1+x)}{x}} {(2-x) + (4-3x)\frac{\log(1+x)}{x}} \quad (0 < x < 1). \end{gather*}

$g(x)$ has the form $\frac{a+by}{c+dy}$, where $a, b, c, d, y$ are functions of $x$, and $b, c, d, y$ are strictly positive. Ignoring the dependence on $x$ for the moment, we see that for $y_, y_\text{max} > 0$, $$ \frac{a+by_\text{max}}{c+dy_\text{max}} - \frac{a+by}{c+dy} = \frac{(bc-ad)(y_\text{max}-y)}{(c+dy_\text{max})(c+dy)}. $$ The denominator is positive, and for the given $a, b, c, d,$ we have: \begin{align*} bc-ad & = (2-x)(4+3x) + (1-2x)(4-3x) \\ & = 8 + 2x - 3x^2 + 4 - 11x + 6x^2 \\ & = 12 - 9x + 3x^2 \\ & = \frac{21}{4} + 3\left(\frac{3}{2} - x\right)^2 > 0. \end{align*} Therefore, if we know that $\frac{\log(1+x)}{x} < y_\text{max}$ for all $x$ in a given interval, then: $$ g(x) < \frac{(2x-1) + (4+3x)y_\text{max}} {(2-x) + (4-3x)y_\text{max}} $$ for those values of $x$. But for all $x > 0$, we have: $$ \frac{d}{dx}\frac{\log(1+x)}{x} = \frac{x - (1+x)\log(1+x)}{x^2(1+x)} < 0, $$ because if $u = -\log(1+x)$, the numerator is $e^{-u}(1 + u) - 1 < 0$; therefore $\frac{\log(1+x)}{x}$ is a strictly decreasing function of $x$. Any lower bound for $x$ therefore gives us a value for $y_\text{max}$. In particular, because $\lim_{x\to0+}\frac{\log(1+x)}{x} = 1$, we can take $y_\text{max} = 1$ for all $x \in (0, 1)$, which gives us: $$ g(x) < \frac{3+5x}{6-4x} \quad (0 < x < 1). $$ Even this simple bound for $g(x)$ is almost enough to prove the required inequality - it just fails, by about $0.05$, for a short interval of values of $x$, about $(0.77, 0.91)$.

Let us use it, at any rate, to prove $f(x) > g(x)$ for all $x \in \left(0, \frac1 2\right]$.

We begin by simplifying the left hand side a little: \begin{equation} \label{eq:3091056:1}\tag{$*$} f(x) > h(x) = \frac{x}{2} + \sum_{n=0}^\infty\frac{x^n}{n+1} = \frac{x}{2} - \frac{\log(1-x)}{x} \quad (0 < x < 1). \end{equation} If $0 < x \leqslant \frac1 2$, then \begin{gather*} g(x) - \frac{x}{2} < \frac{3+5x}{6-4x} - \frac{x}{2} = \frac{3+2x+2x^2}{6-4x} \leqslant \frac{9}{12-8x} \\ = \frac3 4\left(1 - \frac{2x}{3}\right)^{-1} = \frac3 4 + \frac{x}{2} + \sum_{n=2}^\infty a_nx^n, \end{gather*} where $$ a_n = \frac{2^{n-2}}{3^{n-1}} \leqslant \frac{1}{n+1} \quad (n \geqslant 2). $$ Therefore: $$ g(x) - \frac{x}{2} < \sum_{n=0}^\infty\frac{x^n}{n+1} = -\frac{\log(1-x)}{x} \quad \left(0 < x \leqslant \tfrac{1}{2}\right), $$ as required.

On the other hand, if $\frac1 2 < x < 1$, then $$ \frac{\log(1+x)}{x} < 2\log\left(\frac{3}{2}\right) < \frac{9}{11}, $$ so we can take $y_\text{max} = \frac{9}{11}$, which gives us: $$ g(x) < \frac{11(2x-1) + 9(4+3x)}{11(2-x) + 9(4-3x)} = \frac{25+49x}{58-38x} \quad \left(\tfrac{1}{2} \leqslant x < 1\right). $$ This rational function, call it $r(x)$, has the form of a constant term plus a multiple of the convex function $1/(58-38x)$, so it is convex. We use this as a substitute for the convexity of $g(x)$, which is visually obvious, but doesn't look easy to prove.

The lower bound we gave as a convenient substitute for $f(x)$ in \eqref{eq:3091056:1} is also convex, because its derivative increases strictly with $x$: $$ h'(x) = 1 + \sum_{n=1}^\infty\frac{(n+1)x^n}{n+2} \quad (0 < x < 1). $$ We also need the explicit formula: $$ h'(x) = \frac1 2 + \frac1{x(1-x)} + \frac{\log(1-x)}{x^2} \quad (0 < x < 1). $$ Let $t_1(x)$ be the tangent to the graph of $h(x)$ at $x = 0.878$. Numerically, we find $t_1(1) \bumpeq 3.702068 > 3.7 = r(1)$, and $t_1(0.824) \bumpeq 2.451292 > 2.449640 \bumpeq r(0.824)$. Therefore: $$ f(x) > h(x) \geqslant t_1(x) > r(x) > g(x) \quad (0.824 \leqslant x < 1). $$ Let $t_2(x)$ be the tangent to the graph of $h(x)$ at $x = 0.729$. Numerically, we find $t_2(0.824) \bumpeq 2.450471 > r(0.824)$, and $t_2\left(\frac1 2\right) \bumpeq 1.444453 > 1.269231 \bumpeq r\left(\frac1 2\right)$. Therefore: $$ f(x) > h(x) \geqslant t_2(x) > r(x) > g(x) \quad (0.5 \leqslant x \leqslant 0.824). $$ This completes the proof.

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