Prove that $\tanh^2(x) \leq x^2$

Question

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.

Answer 1

Suppose that $x\geqslant 0$. Then$$\sinh x=x+\frac1{3!}x^3+\frac1{5!}x^5+\cdots+\frac1{(2n-1)!}x^{2n-1}+\cdots$$and$$x\cosh x=x+\frac1{2!}x^3+\frac1{4!}x^5+\cdots+\frac1{(2n-2)!}x^{2n-1}.$$Therefore, since we are assuming that $x\geqslant0$, $\sinh x\leqslant x\cosh x$. In other words, $\tanh x\leqslant x$. And, since $\tanh$ is an odd function, $\tanh(-x)=-\tanh(x)\geqslant-x$. So, $(\forall x\in\mathbb R):\bigl\lvert\tanh(x)\bigr\rvert\leqslant\lvert x\rvert$.

Answer 2

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.