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I'm proving Bernoulli's inequality by induction but I noticed something strange.

See wikipedia proof: http://en.wikipedia.org/wiki/Bernoulli's_inequality

Notice how they multiply both sides of the inequality by $(1+x)$ in the 3rd line of the proof.

Now this equals $(1+x)^{k+1}$ on the left side and $(1+x)(1+kx)$ on the right side.

What I prefer to do is to just plug in $(k+1)$ instead of $k$ when I prove by induction (but I am just now learning it so I may be wrong) and therefore I do:

$$(1+x)^k \geq 1+kx, \space \space \space x\geq -1$$ $$=(1+x)^{k+1} \geq 1+(k+1)x$$

But my right hand side is not equal to wikipedia. Why is this?

Cameron Buie
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Paze
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2 Answers2

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You can only conclude that $(1+x)^{k+1}\ge(1+kx).(1+x)$ because you only assume it is true for $k$. But the RHS is equal to $1+(k+1)x+kx^2\ge 1+(k+1)x$ because $kx^2\ge 0$. So it's proved by induction. Note that your method allows us to prove anything.

JefL
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Your induction hypothese here is $$(1+x)^k\geq 1+kx$$ for $x\geq -1$.

Based on that hypothese you are allowed to conclude that: $$(1+x)^{k+1}\geq (1+kx)(1+x)$$ This is also remarked in the answer of JefL. It is not just a question of interchanging $k$ with $k+1$ as you would like to do. If $(1+x)^k\geq 1+kx$ is true then $(1+x)^{k+1}\geq (1+kx)(1+x)$ is definitely true as well, but it is not immediately clear that $(1+x)^{k+1}\geq 1+(k+1)x$ is true also. It is true allright, but there are steps in between to get from: $$(1+x)^{k+1}\geq (1+kx)(1+x)$$ to: $$(1+x)^{k+1}\geq 1+(k+1)x$$which are exposed on the link in your question.

drhab
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  • I can't see how they go from $(1+x)^{k+1}\geq (1+kx)(1+x)$ to $(1+x)^{k+1}\geq 1+(k+1)x$ in my link? They have an added $kx^2$ in their proof. – Paze Jan 23 '14 at 19:36
  • You say 'added' but it is just a question of working out the right side: $\left(1+x\right)\left(1+kx\right)=1+\left(k+1\right)x+kx^{2}$. Do you understand that? Here $kx^{2}\geq0$ leading to $\left(1+x\right)^{k+1}\geq\left(1+x\right)\left(1+kx\right)=1+\left(k+1\right)x+kx^{2}\geq1+\left(k+1\right)x$. Wich is exactly what you want to prove here. – drhab Jan 23 '14 at 19:46