Use the Mean Value Theorem to prove $\cosh(x) \ge 1 + \frac{x^2}{2}$ in the interval $[0,x]$, given $\sinh(x) \ge x$ for all $x \ge 0$.

Use the Mean Value Theorem to prove $\cosh(x) \ge 1 + \frac{x^2}{2}$ in the interval $[0,x]$, given $\sinh(x) \ge x$ for all $x \gt 0$.

I tried using $f(x) = \cosh(x)$, but to no avail. All help appreciated, thanks!

• Are you sure you copied this correctly. It seems that the variable $x$ is used in two different meanings in: prove $\cosh(x) \ge 1 + \frac{x^2}{2}$ in the interval $[0,x]$. (This makes the question unclear. Of course, I might have misunderstood something.) – Martin Sleziak Jun 13 '13 at 12:57
• @MartinSleziak puu.sh/3eJMs.png – Clinton Jun 13 '13 at 12:59

What about the function $f(x)=\cosh (x)-1-\frac{x^2}2$.

You clearly have $f(0)=0$. Can you show $f'(x)\ge0$. If you use these facts and Mean value theorem, what do you get?

Let's assume $\cosh (x) \leq 1 + \frac{x^2}{2}$, within that interval and let $f(x) = \cosh (x)$. Then, the mean value theorem tells us that there exists an $x_0 \in [0,x]$ such that

$$f'(x_0) = \frac{f(x) - f(0)}{x - 0}.$$

If $f(x) = \cosh (x) \implies f'(x) = \sinh (x)$. From the constraint given, we know that $\sinh (x_0) \geq x_0$. Now, let's work out the RHS of the MVT equality:

$$1) \,\,\,\,\,\,\,\,f(x) = \cosh (x) = \cosh (x)$$ $$2) \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f(0) = \cosh (0) = 1$$ $$3) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, x - 0 = x.$$

Subbing this all in, we get:

$$\sinh (x_0) = \frac{\cosh (x) - 1}{x}.$$

As $\sinh (x_0) \geq x_0$ we get

$$x_0 \leq \frac{\cosh (x) - 1}{x}$$ $$x \cdot x_0 \leq \cosh (x) - 1$$ $$\implies \cosh(x) \geq 1 + x \cdot x_0.$$

We know that $x_o \leq x$ and so, the biggest value this $x_0$ can possible take is $x$, putting this in gives us

$$\cosh (x) \leq 1 + x \cdot x = 1 +x^2$$

which is a contradiction as we said that $\cosh (x) \leq 1 + \frac{x^2}{2} < 1 + x^2$, as $x > 0$.

• $\sinh(x)\geq x$ implies $\frac{\cosh(x)-1}{x}\geq x$ – Amr Jun 13 '13 at 12:36
• @Amr Oh yeah, sorry, typo. – Kaish Jun 13 '13 at 12:38
• It seems to me that your argument can be used to show that $\cosh(x)\geq 1+x^2$ (a better lower bound) – Amr Jun 13 '13 at 12:39
• @Amr As opposed to $\geq 1 + \frac{x^2}{2}$? So it doesn't answer this question properly? – Kaish Jun 13 '13 at 12:40
• OK . There is a mistake – Amr Jun 13 '13 at 12:43