Hyperbolic Tangent Less than X

Theorem

$\tanh x \le x$

for $x \ge 0$.

Proof

Let $\map f x = x - \tanh x$.

By Derivative of Hyperbolic Tangent:

$\map {f'} x = 1 - \sech^2 x$

Since $\cosh x \ge 1$ for all $x \in \R$, we can deduce that:

$\map {f'} x \ge 0$

From Derivative of Monotone Function, $\map f x$ is increasing.

By definition of hyperbolic tangent:

$\map f x = 0$

It follows that $\map f x \ge 0$ for $x \ge 0$.

Hence the result.

$\blacksquare$