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Prove $(1+z)^a \geq 1+az$, for $z>-1, a>1$, using the mean value theorem

Hint says try using: $f(z)=(1+z)^a$

I've tried messing around with this, but I can't seem to get 1 + az on the RHS. I'm struggling a little. Would appreciate help on this. Must be some simple trick that I'm missing.

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  • $\begingroup$ To apply the mean value theorem, you need a differentiable function on an open interval. You have the function, as suggested in the hint. Now pick an interval. $\endgroup$ – Raskolnikov Sep 14 '12 at 15:50
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Let $f(z)=(1+z)^a$. By the Mean Value Theorem, $$\frac{f(z)-f(0)}{z-0}=f'(c)$$ for some $c$ between $0$ and $z$. (The number $c$ depends on $z$, and perhaps should be called $c_z$.) Thus $$\frac{(1+z)^a-1}{z-0}=a(1+c)^{a-1},$$ and therefore $$(1+z)^a=1+az(1+c)^{a-1}.\tag{$1$}$$

Now the verification of the inequality splits into two parts. If $z\ge 0$, the right-hand side of $(1)$ is $\ge 1+az$, since $(1+c)^{a-1}\ge 1$.

If $-1\lt z\lt 0$, then since $z$ is negative, and $0\lt (1+c)^{a-1}\lt 1$, we have $1+az(1+c)^{a-1} \gt 1+az$.

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  • $\begingroup$ Thanks makes perfect sense now! :) $\endgroup$ – confused Sep 14 '12 at 16:30
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Define $f(z)=(1+z)^a$.

First case $z=0$: we have equality, Ok.

Second case $z>0$. Mean value theorem: $\frac{f(z)-f(0)}{z}=f'(\eta)$ where $\eta\in]0,z[$. Then observe that $\inf_{\eta\in ]0,z[} f'(\eta)=a$. Then you get the inequality. Complete with the case $z\in]-1,0[$ (Use the $\sup$ of the derivative here).

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