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Thread: [SOLVED] Bernoulli's inequality

  1. #1
    Aug 2008

    Question [SOLVED] Bernoulli's inequality

    I am (somewhat) familiar with proving using induction, but why is proof of Bernoulli's inequality, after assuming validity for P(k), set for P(k+1) as:
    \begin{matrix}<br />
(1+x)(1+x)^k \ge (1+x)(1+kx)<br />
& \iff & (1+x)^{k+1} \ge 1+(k+1)x+kx^2<br />

    I don't understand the right side of inequation, shouldn't it be: \ge 1+(k+1)x

    I see that the inequality must be strictly multiplied by "(1+x)" on both sides (in order to remain the same inequality), but shouldn't one, strictly following rules of mathematical induction, insert "k+1" on both sides, and thus get (1+x)(1+x)^k \ge 1+(k+1)x
    Last edited by courteous; Oct 14th 2008 at 09:39 PM.
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