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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?

5 Answers 5

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

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

which is obviously true.

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    So, does that mean that if $f(x) \geq g(x)$, then $f'(x) \geq g'(x)$? If so, is this a theorem with a name? – 2017-02-22
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    No. $f'-g'$ growing, so $f-g$ stays positive *provided it starts positive*. – 2017-02-22
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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!

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    Isn't it easier to use that the derivative of $\tanh x$ is $\tanh^2-1$? – 2017-02-22
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    @egreg I have edited the answer. Thanks! – 2017-02-22
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HINT: use the analytic definition, that is

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

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

The result follows.

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    Aren't these results harder to establish than the original ? – 2017-02-22
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    @YvesDaoust Yes they are, I suppose as a hint it doesn't do much for the OP as these are hard to prove. For example establishing that $ \cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{(2n+1)^2\pi^2}\right)$ is tricky but I can include the formulation if you feel this would aid the OP. – 2017-02-23
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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 $[0,\infty)$.