I have this inequality $$ \frac{\frac{(1-2 a)x^3}{2}\;\coth \frac{(1-2 a)x^3}{2} }{\frac{x^3}{2}\;\coth \frac{x^3}{2} } <1 , $$ and I want to show that it is valid for $0<a<1$ and $x>0.\;$ To do this, defining $f(y) = y\coth y$, the inequality reduces to
$$ \frac{f(\frac{(1-2 a)x^3}{2}) }{ f(\frac{x^3}{2})} <1 . $$
Now, it is sufficient to show that $f(y)$ is increasing. Calculating $f'(y)$, I obtain $f'(y) = \frac{\sinh 2 y \;-\;2 y}{2 \sinh^2 y}$, which is positive but only for $y>0$, but the argument $\frac{(1-2 a)x^3}{2}$ for $a> \frac{1}{2}$ is negative, which means that I can not use this method, or it needs some further assumptions. If so, I will be grateful if someone suggests other ways that work in this case.
Writing $1-2a = b$, so $-1 < b < 1$, you want to show that $\dfrac{f(bx)}{f(x)} \lt 1$ where $f(x) =x\coth(x) $.
Looking at $f(x) = x\coth(x)$, $f$ is even, $f(x) \ge 1$ and $f(x) > 1$ for $x \ne 0$, $f'(x) > 0$ for $x > 0$ and $f'(x) < 0$ for $x < 0$.
Therefore, if $x \ne 0$ and $|b| < 1$ then, since $bx$ is closer to zero than $x$, $1 \lt f(bx) \lt f(x) $ if $|b| < 1$.