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Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$.

I don't want to use the AM-GM inequality to prove this (from which this statement would follow).

I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by $$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$

Now suppose the statement holds for $n$, then $$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$

So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then $$ x_1 + \dots + x_n x_{n+1} \ge n. $$

Then we also have $$ x_1 + \dots + x_{n-1} + x_n + x_{n+1} \ge n - x_{n}x_{n+1} + x_n + x_{n+1}. $$

Now I'm not sure where else to go from here.

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  • $\begingroup$ Seem similar in proving AM-GM by induction I think $\endgroup$ – Azlif Dec 21 '17 at 23:25
  • $\begingroup$ @MathematicianByMistake Induction certainly works, since this is just a special case of AM-GM which is proven by induction. $\endgroup$ – B. Mehta Dec 21 '17 at 23:29
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Thus, it's enough to prove that $$n-x_nx_{n+1}+x_n+x_{n+1}\geq n+1$$ or $$(x_n-1)(x_{n+1}-1)\leq0,$$ which you can assume before.

Indeed, since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$

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  • $\begingroup$ It's not entirely clear to me why this can be assumed. $\endgroup$ – B. Mehta Dec 21 '17 at 23:31
  • $\begingroup$ @B. Mehta Since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$ $\endgroup$ – Michael Rozenberg Dec 21 '17 at 23:33
  • $\begingroup$ Ah, so there's a relabeling step added earlier. $\endgroup$ – B. Mehta Dec 21 '17 at 23:34
  • $\begingroup$ @B.Mehta I believe the logic is: there has to be at least one value greater-than-or-equal-to 1 and at least one less-than-or-equal-to 1. Rearrange so that these two values come last. $\endgroup$ – Addem Dec 21 '17 at 23:34
  • $\begingroup$ @MichaelRozenberg Thanks, very helpful. I would like to verify this last part with there needing to be a pair with one greater and one and one less. By contradiction, suppose for every $x_i$ and $x_j$, $(x_i -1)(x_j-1) > 0$, then either all pairs are greater than one or less than one which in either case would result in a product of the $x_k$'s being greater than 1 or less than one, respectively. $\endgroup$ – user330531 Dec 21 '17 at 23:42
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The inequality follows from AM-GM, indeed: $\;x_1 + \dots + x_n \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_n}\,$.
Below is a proof of AM-GM by induction, along a different idea than the ones posted already.

Lemma.   $\;(n-1)t^n+1 \ge n t^{n-1}\;$ for all $\;t \ge 1\,$.

Let $f\,(t)=(n-1)t^n - n t^{n-1} + 1\,$, then $\,f(1) = 0\,$ and $\,f'(t)=n(n-1)t^{n-2}(t-1) \ge 0\,$, so $\,f\,$ is increasing from $\,f(1)=0\,$ for $\,t \ge 1\,$ and therefore $\,f(t) \ge 0\,$ for $\,t \ge 1\,$.

(The stronger statement is true that $\,f(t) \ge 0\,$ for all $\,t \ge 0\,$. That follows from the identity $\,(n-1)t^n - n t^{n-1} + 1 = (t-1)^2 \cdot \sum_{k=1}^{n-1} k \cdot t^{k-1}\,$ which has an elementary no-calculus proof.)

Induction step.   The AM-GM inequality is homogeneous, so it can be assumed WLOG that the smallest $x_j=1\,$. Assume again WLOG that $\,j=n\,$, then the inequality reduces to:

$$x_1+x_2+ \cdots + x_{n-1} + 1 \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}} \quad\style{font-family:inherit}{\text{with}}\quad x_i \ge 1, \;i=1,2,\ldots,n-1$$

By the induction hypothesis:

$$\,x_1+\cdots+x_{n-1} \ge (n-1) \sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} \tag{1}$$

Also, it follows from the lemma with $\,t = \sqrt[n(n-1)]{x_1\cdots x_{n-1}}\ge 1\,$ that: $$\,(n-1)\cdot\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \ge n \cdot \sqrt[n]{x_1\cdot \ldots \cdot x_{n-1}} \tag{2}$$

Therefore:

$$ x_1+ \ldots +x_{n-1}+1 \;\stackrel{(1)}{\ge}\; (n-1)\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \;\stackrel{(2)}{\ge}\; n \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}} $$

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