<s>[INST] Mean Value Theorem to show inequalities about numbers 
Show $\sqrt{65}-8=\frac{1}{2\sqrt{c}}$ for some $c \in (64,65)$, and hence show: 
  $$8+\frac{1}{18}<\sqrt{65}<8+\frac{1}{16}$$

I managed to do the first part easily, but I don't know how to do the "hence show" part. 
Working For the First Part:
Consider the function $f(x)=x^{\frac{1}{2}}$ on $[64,65]$. Then f is continuous on $[64,65]$ and differentiable on $(64,65)$. By the mean value theorem, there is a $c \in (64,65)$ such that:
$$f'(c)=\frac{f(65)-f(64)}{65-64}$$
$f'(x)=\frac{1}{2\sqrt{x}}$, so we have:
$$\frac{1}{2\sqrt{c}}=\sqrt{65}-8$$
 [/INST] The first part looks good.
From the first part we know, that there exists a $c \in (64, 65)$, such that 
$$\sqrt{65} = 8 + \frac{1}{2\sqrt{c}} \; .$$
Since $64 < c < 81$ and the square root is a monotone function, we see that $$8 < \sqrt{c} < 9 \; .$$ 
From this it follows that 
$$ \frac{1}{18} = \frac{1}{2\cdot 9} < \frac{1}{2\sqrt{c}} < \frac{1}{2\cdot 8} = \frac{1}{16} \; ,$$
so by adding $8$ we get 
$$ 8 + \frac{1}{18} < \underbrace{8 + \frac{1}{2\sqrt{c}}}_{= \sqrt{65}} < 8 + \frac{1}{16} \; .$$
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