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In Cauchy's forward-backward induction proof, why can we substitute $x_k=\frac{x_1+x_2+ ... +x_{k-1}}{k-1}$ without losing generality?

Michael Rozenberg
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Shafin Ahmed
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1 Answers1

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We can not substitute it of course in the general case, but if we'll prove that for $P(k): \frac{x_1+x_2+...+x_k}{k}\geq\sqrt[k]{x_1x_2...x_k}$ and fo $x_i>0$ the following statements are true: $$P(2)$$ $$P(k)\Rightarrow P(2k)$$ and $$P(k)\Rightarrow P(k-1)$$ then it's obvious that we'll prove AM-GM.

For the proof of the last statement we can assume $x_k=\frac{x_1+x_2+...+x_{k-1}}{k-1}$ because we assumed that $P(k)$ is true for all positive $k$ numbers $x_i$ and from here

it's true for $k$ positive numbers $x_1$,...,$x_{k-1}$,$\frac{x_1+x_2+...+x_{k-1}}{k-1}$, which gives that $P(k-1)$ is true.

Michael Rozenberg
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  • This way aren't we proving $P(k)\Rightarrow P(k-1)$ for some particular sequence of $k$ positive numbers rather than all sequence of $k$ positive numbers? – Shafin Ahmed Jun 21 '19 at 16:48
  • $P(k)\rightarrow P(k-1)$ it's: with assuming that AM-GM is true for all positives $x_1$, $x_2$,...,$x_k$ we need to prove that AM-GM is true for all positives $x_1$,..., $x_{k-1}.$ The proof of this statement you saw already. – Michael Rozenberg Jun 21 '19 at 16:59