Prove an Inequality with $2$ Different Logs

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.

• What is $q$ in the numerator? – Shubham Johri Jan 28 '19 at 16:10
• @ShubhamJohri it is supposed to be $x$. Corrected. – Bill.Math Jan 28 '19 at 16:12
• The Taylor approximation helps, but we need to work with polynomial of twenty degree. It's true for all $0<x<1$. – Michael Rozenberg Jan 28 '19 at 17:57
• How do you make out that $f_2(x)\to\infty$ as $x\to1$? – Calum Gilhooley Jan 28 '19 at 19:24
• @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. – Bill.Math Jan 28 '19 at 20:17

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: $$$$\label{eq:3091056:1}\tag{*} f(x) > h(x) = \frac{x}{2} + \sum_{n=0}^\infty\frac{x^n}{n+1} = \frac{x}{2} - \frac{\log(1-x)}{x} \quad (0 < x < 1).$$$$ If $$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.