How can I prove that $\sinh (x) \leq x \cosh (x)$ for all $x \geq 0$?

Question

I'm currently struggling to prove that, for all $x \geq 0$, $\sinh (x) \leq x \cosh (x)$. Or equivalently, that

$$ \frac{e^x - e^{-x}}{2} \leq \frac{xe^x + xe^{-x}}{2} \hspace{5 mm} \Rightarrow \hspace{5 mm} (1-x)(e^x - e^{-x}) \leq 0 $$

I've attempted to prove this by induction, but have have been unable to use the induction hypothesis $x=k$ to prove that the case $x=k+1$ must hold.

I've also attempted to prove this by contradiction, by supposing that

$$ (1-x)(e^x - e^{-x}) > 0 $$

but have been unable to reach a contradiction.

Can anyone advise me how to go about tackling this problem?

Answer 1

$$\sinh(0)\le0\cosh(0)$$ is true, and by differentiation

$$\cosh(x)\le\cosh(x)+x\sinh(x)$$

which is obviously true.

Answer 2

HINT: use the analytic definition, that is

$$\sinh x=\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!},\quad \cosh x=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}$$

Answer 3

Dividing each side by $\cosh x$, we have to prove that $$x-\tanh x \ge 0$$ For $x \ge 0$.

However, the derivative of $x-\tanh x$ is $$\tanh^2 x \ge 0$$ So, $x-\tanh x$ is an increasing function. However, $0-\tanh 0=0$. So $x \ge \tanh x $.

Done!

Answer 4

Hint: Since $$\cosh x \leq e^{x^{2}/2}$$ and $$\frac{\sinh x}{x} \leq e^{x^{2}/6} $$

The result follows.

Answer 5

It is sufficient to consider when $x > 0$. By the mean value theorem, there is $\xi \in (0, x)$ such that

$$ \frac{\sinh x}{x} = (\sinh)'(\xi) = \cosh(\xi) < \cosh x $$

since $\cosh$ is incresing on $[0,\infty)$.