Hyperbolic Tangent Less than X

Theorem

 * $\tanh x \le x$

for $x \ge 0$.

Proof
Let $f \left({x}\right) = x - \tanh x$.

By Derivative of Hyperbolic Tangent Function, $f' \left({x}\right) = 1 - \operatorname{sech}^2 x$.

Since $\cosh x \ge 1$ for all $x \in \R$, we can deduce that $f' \left({x}\right) \ge 0$.

From Derivative of Monotone Function, $f \left({x}\right)$ is increasing.

By Definition of Hyperbolic Tangent, $f \left({x}\right) = 0$.

It follows that $f \left({x}\right) \ge 0$ for $x \ge 0$.