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#1




Proof that (1+x)^n >= 1+nx?
Prove that (1+x)^n >= 1+nx for x>1 and n any natural number.
I just got done with a math test*, and I could not for the life of me figure out the above. I was totally stumped. I rewrote it by defining y:=x+1 to obtain y^n >= 1+n(y1) for y>0. I then noticed that for the trivial case where n = 1, the statement is y >= 1+1(y1) which is obviously true. Also, when y=1, the statement is 1^n >= 1+n(11) which is also true. Aside from these obvious observations, I made no more headway. I assume that a proof by contradiction or a proof by induction is possible, as those two techniques were what the test was about. If anyone out there could give a proof, or an outline of a proof, or some helpful advice, I'd be grateful. * Meaning that this is not a homework question, and I'm only asking so that I may gain knowledge. 
#2




Since it is impossible to tell the real reason why someone asks such homeworktype questions on "Teh Intranets", I only give hints. (One would think this would be SDMB policy but it isn't.)
You didn't happen to be studying the Binomial Theorem did you? 
#3




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Try starting with the assumption that it's true for n and multiply both sides of the inequality by (1+x) (which is positive if x > 1) to get (1+x)^{n+1} >= ... something which can be shown to be >= 1 + (n+1)x. I'll let you work out the details. 
#4




The OP is incomplete, because it has to be about the limit as x approaches something. (And those that have responded seem to be assuming that x is approaching 0, which is probably the right assumption).



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#7




Following Thudlow: Restrict x > 1, so 1+x > 0.
 For n = 1, obviously (1+x)^{1} = (1+x) = 1 + nx.  Suppose true for n. Then (1+x)^{n+1} = (1+x)^{n}(1+x) >= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0 = 1 + nx + x + nx^{2} by multiplication >= 1 + nx + x since x^{2} >= 0 = 1 + (n+1)x QED. Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0 
#8




You can also prove it using calculus, though I leave it as an exercise to the reader (snerk).

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#12




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1 + nx + x + nx^{2} >= 1 + nx + x Since n and x^{2} are both nonnegative, the left side consists of the same thing as what's on the right, but with something nonnegative added to it. 
#13




Funnily enough, I just used that example in the exercises yesterday to discuss proof by induction. Otherwise, everything of substance has already been said in this thread, so all that's left for me is to point out that the inequality in the OP is generally known as Bernoulli's inequality.

#14




This is a very practical question, actually, since it amounts to answering the question of why compound interest is better than simple interest.



#15




OP....
You have not included Z^%√7 Later, Jaxon.. 
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Lest anyone not realize, this is a zombie thread. Not that there's any harm in that.

#17




I wonder if the homework ever got turned in?
(I love zombie threads in October) 
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Note to self... Do.Not.Read.'Math'.Threads.
__________________
"Always do sober what you said you'd do drunk. That will teach you to keep your mouth shut." Ernest Hemingway (1899  1961) 
#19




For what it's worth, we can also splay out the argument like this:
Suppose n >= 1 and b >= 0. Then (b  1)^{2} * [1 + (1 + b) + (1 + b + b^{2}) + ... + (1 + b + b^{2} + ... + b^{(n  2)})], being a square times a sum of terms each >= 0, is >= 0 as well. Expanding this out and cancelling and collecting terms, we get (n  1)  nb + b^{n} = b^{n}  (1 + n(b  1)). Thus, we've shown that b^{n} >= 1 + n(b  1) for b >= 0, which is to say, (1 + x)^{n} >= 1 + nx for x >= 1, as desired. [This is essentially the same calculation as powers the inductive argument, just set down differently. (Incidentally, the calculus argument is also just a slight reframing of the same reasoning as in the inductive argument or this one)] 
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