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I want to show that $(1-r)^x \ge (1-xr)\ \forall x \in \mathbb{N}, r \in \mathbb{R}, 0 \le r \le 1$

This is what I have so far, I'm hoping for a critique as well as help with the end since I don't think I'm quite there yet.

Base:

LHS =$(1-r)^1 = (1-r)$

RHS = $(1-1r) = (1-r)$

And the base case holds.

Hypothesis: Assume $(1-r)^k \ge (1-kr)$ is true for some $k \in \mathbb{N}$

Conclusion: RHS = $1-(k+1)r$

LHS: \begin{align*} (1-r)^{k+1}&=(1-r)^k(1-r)\\ & \ge (1-kr)(1-r) \quad \textrm{From Hypothesis}\\ &= 1-r-kr+kr^2\\ &= 1-kr-r+kr^2\\ &=1-(k+1)r+kr^2\\ \end{align*}

So, from here it seems clear to me that since $k\in \mathbb{N}$ and $r^2$ is a perfect square, and therefore $\ge 0$, that my LHS $\ge$ RHS, and this would prove that it's true. But - I'm not sure that my reasoning is sound, or if I need to show more?

Edit to add: Is regular induction enough for this? Are there values that I haven't accounted for?

Math1
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    Yes it's correct;Nothing more to show – Learnmore Dec 01 '16 at 03:31
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    The proof is OK. Two minor comments: (1) starting with $x=0$ is even simpler and gives you a stronger conclusion, namely that the inequality holds for all natural numbers, including $0$. (2) $r^2$ is a square; perfect square is reserved for the squares of integers. – Fabio Somenzi Dec 01 '16 at 03:34
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    So, could it simply say $r^2$ is a square, therefore it is $\ge 0$?...Same idea as a perfect square except not an integer? – Math1 Dec 01 '16 at 03:38

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