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How can I show that:

$$ a^n-1 \geq n\left(a^{\frac{n+1}{2}}-a^\frac{n-1}{2}\right)$$

$$ \sum_{k=0}^{n-1} a^k \geq na^\frac{n-1}{2}$$

$$ a>1, n\in\mathbb{N} $$

without studying the function

$$ f(x)=x^n-1 - n\left(x^{\frac{n+1}{2}}-x^\frac{n-1}{2}\right)$$?

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1 Answer 1

up vote 6 down vote accepted

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+..+a+1)$$

By AM-GM

$$a^{n-1}+a^{n-2}+..+a+1 > na^{\frac{n-1}{2}}$$

Thus

$$a^n-1 >(a-1)na^{\frac{n-1}{2}}$$

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Thanks for noting me that. I misread the title. :) –  B. S. Aug 28 '12 at 13:17
    
Thank you for your answer! –  Chon Aug 28 '12 at 16:19
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