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!
 A: 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?
A: 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 $x > 0$.
