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Using Mean Value Theorem, prove that:

$$\sqrt{1+h}<1+\frac{1}{2}h \text{ for } h>0.$$

Now, generalize this as follows: If $0<p<1$ and $h>0,$ then show that

$$(1+h)^p<1+ph.$$

Assume the usual rules about differentiating powers.

I have no idea, how to even start this. I can state the mean value theorem, but it's hard to see how that's used as there is no explicit function stated.

MVT: If $f:[a,b]→\mathbb R$ is continuous on $[a,b]$ and differentiable on $(a,b)$, Then there is a $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$

I suppose I could generate functions, say, $f(x)=\sqrt{1+x}, g(x)=1+\frac{1}{2}x, m(x)=\sqrt{1+x}-(1+\frac{1}{2}x)=f(x)-g(x)$. I want to show that $m(x)<0$ for $x>0$?

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  • $\begingroup$ Consider $f(x)=x^p$ and estimate the derivative. $\endgroup$ – lzralbu Nov 12 '19 at 20:02
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Hint:

Just consider

$$\frac{(1+h)^p-1}{h} = \frac{p}{(1+\xi)^{1-p}} < p$$

where $0<\xi<h$.

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You can use $f$, where taking $x=h$ and $x=0$, for the mean value th. you have that exists $c\in (0,h)$ such that $\sqrt{1+h} - 1 = f(h)-f(0) = hf'(c)$. The next step is to dimension $f'(c)$, you do it!

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