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In the book Concrete Mathematics, exercise 1.9, one is given:

$P(n): x_1...x_n \leq (\frac{x_1 + ... + x_n}{n})^n$ if $x_1, ..., x_n \geq 0$.

and is asked the following question:

By setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$, prove that $P(n)$ implies $P(n-1)$ whenever $n > 1$.

The solution to this is given here: Backwards induction to show that $x_1\cdots x_n \leq ((x_1+\cdots+x_n)/n )^n$, but my question is:

Why are we setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$? This equation isn't true for all $x_i$ and all $n$, since taking $x_1 = 0$, $x_2 = 1$ and $n = 2$, the above equation results in $x_2 = x_1$. Aren't we trying to prove $P(n)$ in general, so for all all $x_i$ and all $n$?

EDIT: to be clear, my question is not on exercise 1.9.b, nor on 1.9.c, but as stated, on 1.9.a. More precisely, I don't understand how one is allowed to conclude the general statement of $P(n)$ implies $P(n-1)$ by choosing a particular value of $x_n$.

J. Schmidt
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1 Answers1

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You have left out an important part. The point of the exercise is that it is being shown that $P(n)\implies P(2n)$. This shows $P(2^n)$ for $n\in\mathbb{N}$, but misses $P(k)$ for $2^{n-1}\lt k\lt2^n$. Showing $P(k)\implies P(k-1)$ fills this gap.

From the Comments

We are given $P(n)$. This works for all non-negative values of $x_1,x_2,x_3,\dots,x_n$. We are choosing $x_n=(x_1+x_2+x_3+\dots+x_{n-1})/(n-1)$. Apply $P(n)$ which works for all values of $x_1,x_2,x_3,\dots,x_n$. Then we see that this implies $P(n-1)$.

robjohn
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  • This answer uses a similar strategy. Perhaps reading it might shed light on this. – robjohn Mar 03 '21 at 09:39
  • I don't think this answers my first question. $P(k) \implies P(k-1)$ doesn't hold for all values, only if $x_k = (x_1 + ... + x_{k-1}) / (k-1)$. – J. Schmidt Mar 03 '21 at 09:41
  • Read part (a) of the accepted answer. It only assumes $P(n)$, and shows $P(n-1)$. Then if we know $P(2^n)$, we know $P(k)$ for all $k\le2^n$. – robjohn Mar 03 '21 at 09:42
  • That is what I don't understand: you say that the backwards induction allows us to go from $2^n$ to $2^n-1$, \textit{etc.} I don't agree with this, since we're explicitly setting $x_n = (x_1 + ... + x_{n-1}) / (n-1)$ for the backwards induction. – J. Schmidt Mar 03 '21 at 09:46
  • So? Where is the problem with that? we are not assuming that it is greater than $(x_1x_2\dots x_{n-1})^{1/(n-1)}$, we are showing that, so there is no problem. – robjohn Mar 03 '21 at 09:49
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    We are given $P(n)$. This works for all values of $x_1,x_2,x_3,\dots,x_n$. We are choosing $x_n=(x_1+x_2+x_3+\dots+x_{n-1})/(n-1)$. Apply $P(n)$ which works for all values of $x_1,x_2,x_3,\dots,x_n$. Then show that this implies $P(n-1)$. – robjohn Mar 03 '21 at 09:56
  • Ooh, ok I think I understand by your explanation just now. To paraphrase, setting $x_n$ to a particular value doesn't change the fact that the other $x_i$ can be any value, hence proving $P(n-1)$ in general? I can accept your answer if you replace it with your last comment, or I can adapt it if you like. Thanks! – J. Schmidt Mar 03 '21 at 10:01
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    I have already added my last comment as that point should not be left in comments only. – robjohn Mar 03 '21 at 10:04