Prove that $\tanh^2(x) \leq x^2$ I'm looking for an alternative proof to prove that $\tanh^2(x)\leq x^2$. My current proof is by observing that $\tanh'(x)=1-\tanh^2(x)\leq 1$, hence integrating for positive $x$ gives that $\tanh(x)\leq x$ hence $\tanh^2(x)\leq x^2$, the proof follows by parity of both $\tanh^2$ and $x^2$.
Ideally I wouldn't want the use of derivatives or parity.

Using the identity $\tanh(x) = \frac{e^{2x}-1}{e^{2x}+1}$, we can verify that
$$ \tanh(x) = \frac{2\tanh(x/2)}{1+\tanh^2(x/2)} $$
and hence
$$ \tanh^2(x) \leq (2 \tanh(x/2))^2. $$
Applying this inequality repeatedly, we get
$$ \tanh^2(x) \leq (2^n \tanh(x/2^n))^2. $$
Invoking the fact that $\frac{\tanh t}{t} \to 1$ as $t \to 0$, we get $2^n \tanh(x/2^n) \to x$ as $n\to\infty$ and hence the desired inequality follows.
