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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 Edit

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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Binomial inequality

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