I want to prove this inequality $$\coth (x)-(1-y) \coth [x (1-y)]<\frac{4 x}{1-x^4},$$ where for $0<y<2$ and $x>0$. I tried to simplify the inequality to use the monotonicity of the function on one side, but the relations for first and second derivatives were much more complicated than the original one.
Does anybody have an idea to do it in an easier way?
For $x > 0,$ the left hand side of the proposed inequality tends to $0$ as $y \to 0,$ so the inequality cannot hold if $x > 1,$ and of course it makes no sense if $x = 1.$ We must suppose, therefore, that $0 < x < 1.$
If $y = 1 \pm t,$ where $t \neq 0,$ then what has to be proved is: \begin{equation} \label{3808663:eq:1}\tag{$1$} \coth x - t\coth(tx) < \frac{4x}{1 - x^4} \quad (0 < x < 1, \ 0 < t < 1). \end{equation}
We have $$ \lim_{t \to 0}(t\coth(tx)) = \lim_{t \to 0}\frac{2t}{e^{2tx} - 1} = \frac1x, $$ so when $y = 1,$ although \eqref{3808663:eq:1} does not strictly make sense, it is reasonable to interpret it as: $$ \coth x - \frac1x < \frac{4x}{1 - x^4}. $$
In fact (this will be proved later): \begin{equation} \label{3808663:eq:2}\tag{$2$} \frac1u < \coth u < u + \frac1u \quad (0 < u < 1). \end{equation} Using both parts of \eqref{3808663:eq:2}, we deduce: \begin{equation} \label{3808663:eq:3}\tag{$3$} \coth x - t\coth(tx) < \coth x - \frac1x < x \quad (0 < x < 1, \ 0 < t < 1). \end{equation} This inequality \eqref{3808663:eq:3} is much stronger than the required inequality \eqref{3808663:eq:1}, because of course: $$ x < 4x < \frac{4x}{1 - x^4} \quad (0 < x < 1). $$
It remains to prove \eqref{3808663:eq:2}. The second part is proved here: Is this inequality true? $\coth x\leq x^{-1}+x$. The proof is short, so I'll copy it out:
From the series expansion of the exponential function, if $u > 0$, then: $$ \frac{e^{2u}-1}2 > u + u^2, $$ therefore: $$ \coth u = 1 + \frac2{e^{2u}-1} < 1 + \frac1{u + u^2} = 1 + \frac1u - \frac1{1 + u} = \frac{u}{1 + u} + \frac1{u} < u + \frac1u. $$
For the first part of \eqref{3808663:eq:2}, we have: \begin{align*} \coth u > \frac1u & \iff \frac2{e^{2u}-1} > \frac1u - 1 \\ & \iff \frac{e^{2u} - 1}2 < \frac{u}{1 - u} \\ & \iff e^{2u} < \frac{1 + u}{1 - u} \quad (0 < u < 1). \end{align*} The last inequality is proved in several ways here: prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$ . I'll copy out a short proof I gave here:
The curve $y = \tfrac{1}{x}$ lies above the tangent at $(1, 1),$ therefore: $$ 2u < \int_{1-u}^{1+u}\frac{dx}{x} = \log\frac{1 + u}{1 - u} \quad (0 < u < 1). $$
This completes the proof of \eqref{3808663:eq:2}, therefore \eqref{3808663:eq:3}, therefore \eqref{3808663:eq:1}.
