After simplifying an equation I got this: $$ n*(\frac{y}{x})^{n-1}<1+\frac{y}{x}+(\frac{y}{x})^{2}+...+(\frac{y}{x})^{n-1}<n $$ While $$ n\geq2 $$ $$ x>y\geq0 $$ While n in N and x,y in Q.
It seems obvious that the right side is bigger than the middle, but how do I write a proof for that? Which method do I use both for proving the left inequation and the right one?
On
Let $n \in \mathbb{N}$ and $x,y \in \mathbb{Q}$ such that : $n\geq 2$ and $x>y\geq 0$
Then $\forall k \in \left\lbrace1,....,n-1\right\rbrace$ : $$\dfrac{y}{x}^{n-1} < \dfrac{y}{x}^k < 1$$
So $$\displaystyle \sum_{k=0}^{n-1} \dfrac{y}{x}^{n-1} < \sum_{k=0}^{n-1} \dfrac{y}{x}^k < \sum_{k=0}^{n-1} 1$$
Finally : $$\displaystyle n\dfrac{y}{x}^{n-1} < \sum_{k=0}^{n-1} \dfrac{y}{x}^k < n$$
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You want to show $n*(\frac{y}{x})^{n-1}<1+\frac{y}{x}+(\frac{y}{x})^{2}+...+(\frac{y}{x})^{n-1}<n $.
Let $\frac{y}{x} = r$, so $0 < r < 1$. Your statement becomes
$nr^{n-1} \lt \sum_{k=0}^{n-1}r^k \lt n $.
Since $0 < r < 1$, $r^j < r^k$ when $j > k$ so, for $n \ge 2$, $\sum_{k=0}^{n-1}r^k =r^{n-1}+\sum_{k=0}^{n-2}r^k \gt r^{n-1}+\sum_{k=0}^{n-2}r^{n-1} = n r^{n-1} $.
The right side is even simpler. Since $0 < r < 1$, we have $0 < r^k < 1$ for $k \ge 1$ so $\sum_{k=0}^{n-1}r^k \lt \sum_{k=0}^{n-1}1 =n $.
Since $x>y$, we know that $\frac{x}y < 1$ (divide both sides by $y$). After multiplying with $\left(\frac{x}y\right)^{n-1}$, it then becomes obvious that $\left(\frac{x}y\right)^n < \left(\frac{x}y\right)^{n-1}$ for an arbitrary natural $n$. An inductive argument using the previous results will even yield $\left(\frac{x}y\right)^n < 1$ (for all $n > 0$).
We can use this to prove the right inequality in your question: $$1 + \left(\frac{x}y\right) + \left(\frac{x}y\right)^2 + \dots + \left(\frac{x}y\right)^{n-1} < 1+1+\dots+1 = n.$$ A similar argument will prove the left inequality.
I don't see a nice way to prove both inequalities at once. I think you're better off proving them seperately.