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The binomial inequality is a statement about raising a sum to a natural power:

$ \forall x \in \R, x \ge -1, $ and $ \forall n \in \N : (1 + x)^n \ge 1 + nx. $

It is essentially a truncated version of the binomial theorem, that is easily proved by induction.

Proof

Check it's true for $ n = 1 : (1 + x) \ge 1 + x, $ clearly.

Now assume true for some $ n = k \in \N : (1 + x)^k \ge 1 + kx. $

As $ 1 + x \ge 0, $ we can multiply that inequality through by $ (1 + x) $ and preserve the inequality:

$ (1 + x)^{k+1} \ge (1 + x)(1 + kx) = 1 + x + kx + kx^2. $

As $ kx^2 \ge 0 \forall x \in \R, \forall n \in \N : 1 + x + kx + kx^2 \ge 1 + (1 + k)x. $

So, $ (1 + x)^{k+1} \ge 1 + (k + 1)x, $ so by induction it is true for all $ n \in \N. $

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