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