Let a,b,c be given real numbers such that a > b > c >0. Prove that there exists a number N such that:
a^n > b^n + c^n for every n>= N
a^n > b^n + c^n for every n>= N
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Ok so the right hand side, b^n+c^n, is less than b^n+b^n = 2b^n since c^n < b^n for any positive n. That said all we have to do is show there exists a number N such that a^n > 2b^n for every n >= N. To do this we can write a = br for some r > 1. Then a^n = (br)^n = (b^n)(r^n). Substituting this into the inequality above, we wish to show there exists N such that (b^n)(r^n) > 2b^n for every n >= N. Dividing through by b^n, this is true if and only if there exists N such that r^n > 2 for every n >= N. At this point all that's left is to show r^n > 2 for sufficiently large n, which is true and can be proven, but I think I'll stop here.
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That looks like Fermat's Last Theorem.
http://en.wikipedia.org/wiki/Fermat%27s_…
http://en.wikipedia.org/wiki/Fermat%27s_…
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