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Mathworld has the generalized Bernoulli's inequality $$(1+x)^a>1+ax,$$ which holds for nonzero $x>-1$ if $a>1$ or $a<0$.

I have seen the classical proof done by induction when the exponent is an integer, but I have not seen a proof for this general case. I have a hunch that this involves a Taylor series expansion since $$(1+x)^a=1+ax+\sum_{n=2}^{\infty}\binom{a}{n}x^n.$$ The sum is alternating so it would involve some work showing that it is nonnegative. However, I overlooked the fact that the Taylor expansion about zero will only hold for $|x|<1$.

Is there a way I can overstep this issue or do I need a new approach?

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    $\begingroup$ For $x\ge 0$, and $a\gt 1$, let $f(x)=(1+x)^a-(1+ax)$. Note that $f(0)=0$, and $f'(x)=a(1+x)^{a-1}-a\gt 0$, so $f$ is increasing. $\endgroup$ – André Nicolas Aug 1 '16 at 1:26
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Consider $f(x) = (1+x)^a - 1 -ax\implies f'(x) = a(1+x)^{a-1}-a= a((1+x)^{a-1}-1)$. We have some cases to look at:

$1)$ $a > 1$: If $-1 < x < 0 \implies (1+x)^{a-1} < 1 \implies f'(x) < 0\implies f(x) > f(0) = 0$ . If $x > 0 \implies (1+x)^{a-1} > 1 \implies f'(x) > f(0) = 0$. If $x = 0, f(0) = 0$. Thus equality occurs.

$2)$ $a < 0$: If $-1 <x < 0 \implies (1+x)^{a-1} > 1 \implies f'(x) < 0\implies f(x) > f(0) = 0$. If $x > 0\implies (1+x)^{a-1} < 1\implies f'(x) > 0\implies f(x) > f(0) = 0$. If $x = 0, f(0) = 0$. Thus equality also occurs.

In either case, you have $f(x) \ge 0$.

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