How to prove that for every real number $x\geq0$ and every integer $n\geq0$ $$1+(n+1)x \le (1+x)^{n+1}.$$
I'm trying with the geometric mean and arithmetic mean inequality, but I don't know what $x_1, x_2, ... $ to get to this.
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sorry, I forgot to mention that $x_1, x_2, ...$ are positive! – just1frustredstudent Oct 16 '20 at 12:55
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Do you know the generalized formula for (a+b)^n? Try to factor it out and see what cancels – Nurator Oct 16 '20 at 12:55
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ooh, I see, thank you! – just1frustredstudent Oct 16 '20 at 12:56
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1This is known as Bernouilli's inequality – Arnaud D. Oct 16 '20 at 12:58
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https://en.wikipedia.org/wiki/Bernoulli%27s_inequality#:~:text=In%20mathematics%2C%20Bernoulli's%20inequality%20(named,often%20employed%20in%20real%20analysis.&text=Bernoulli's%20inequality%20 – Albus Dumbledore Oct 16 '20 at 13:00
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See also https://math.stackexchange.com/questions/475309/proof-by-induction-of-bernoullis-inequality-1-xn-geq-1-nx – Arnaud D. Oct 16 '20 at 13:03
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If $x\geq0$ then by the binomial theorem $$(1+x)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}x^n\geq\binom{n+1}{0}x^0+\binom{n+1}{1}x^1=1+(n+1)x.$$ If we allow $x<0$ then there exist counterexamples, such as $x=-4$, $n=2$.
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