Is it possible to prove this inequality with a geometric sum

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I am wondering if the following inequality is true:

$$\sum_{k=0}^{N-1}x^k \leq Nx^N \qquad \forall x\geq 1$$

I have verified the inequality for different values of $N$ but I am looking for a way to prove that it holds for all $N\in\mathbb{N}$.

Any help is appreciated in either proving this or showing that it is not true. Thanks in advance!

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This is true if $x\geq 1$. To see this, expand the r.h.s. as $$Nx^N = x^N+x^N+\dots+x^N$$ where the sum has $N$ terms. Now, couple each one of these with the terms at your left side, you'll have $$x^k\leq x^N$$ for every $k<N$ since $x\geq 1$. Now you sum up these inequalities to get your result.

Other way to do it is using the formula for geometric sums. Thus your inequality is $$\frac{x^N-1}{x-1} \leq Nx^N$$ and I'm pretty sure you can handle from here.