For $t\in(1,\infty)$, the equation $$\cosh(tx)=t\cosh(x)\tag{1}$$ defines $x(t)>0$.Show that $$\coth(tx)<t \tanh(x)\tag{2}$$ for such $x$. I have tried using convexity of $\cosh,\sinh$ to obtain inequalities like $$\cosh\left[\left(t+\dfrac{1}{t}-1\right)x\right]\lt\left(t+\dfrac{1}{t}-1\right)\cosh x$$ $$t\sinh\left(x\right)\lt\sinh\left(tx\right)$$etc without making much headway. How does one show (2)? A less strict inequality $\coth(tx)<tx$ also solves my purpose.
Thanks