2

There are 2011 positive numbers with both their sum and the sum of their reciprocals equal to 2012. Let $x$ be one of these numbers. Find the maximum value of $x + \frac{1}{x}.$

doingmath
  • 419
  • 1
    We know that, for arbitrary positive number $x$, we have $x + \frac{1}{x} \ge 2$. So the sum of both $2011$ positive numbers and their reciprocals is equal or greater than $2011\times2=4022$, it cannot be equal to $2012$. You may double-check your question. – Zeta Dec 24 '19 at 03:05
  • 2
    @Zeta, I think the OP means that the sum of the numbers is equal to $2012$ and so is the sum of their reciprocals (so together they total $4024$, which exceeds $4022$, as you observe they must). – Barry Cipra Dec 24 '19 at 03:41

1 Answers1

1

Consider that there are $2010$ numbers, $y_1$ to $y_{2010}$, which are not equal to $x$. This means that the sum of $y_1$ to $y_{2010}$ equals to $2012-x$ and the sum of their reciprocals is equal to $2012-\frac1x$. Given the Cauchy Schwarz inequality, which states :

enter image description here

You get that :

enter image description here

And therefore:

enter image description here

Which means that your answer to this question is $\frac{8045}{2012}$. This question is a great use of inequalities. You can find more inequalities and applications for these inequalities at : https://brilliant.org/wiki/classical-inequalities/#cauchy-schwarz-inequality. I apologies my use of images as I am not yet very familiar with MathJax. You can find pretty much the exact same solution at https://sumo.stanford.edu/pdfs/smt2011/algebra-solutions.pdf as I have just reworded it a bit.

JC12
  • 1,040
  • This establishes a bound, but the question asks for the maximum. How do you know that the bound can be attained? – joriki Dec 24 '19 at 04:53
  • In inequalities, the bound should be the maximum in this situation. The use of inequalities means that the bound can either be the minimum or maximum. – JC12 Dec 24 '19 at 05:25
  • How do you mean, it "should be"? I think it is, but I don't see an argument for that in your answer. In the Cauchy-Schwarz inequality, equality is attained if and only if the one vector is a multiple of the other. So you need to show that things work out if you make all the $y_i$ equal – that's the only way you can fulfil that condition. – joriki Dec 24 '19 at 05:47
  • You rearrange the second equation to get the third equation and hence the inequality. I don’t understand what you’re trying to say. – JC12 Dec 24 '19 at 06:04
  • I don't question the inequality. I started my first comment by acknowledging that you established a bound. $x+\frac1x$ can't be more than $\frac{8045}{2012}$. But a maximum is not just an upper bound; it's the highest value actually attained. So to prove that $\frac{8045}{2012}$ is the maximum, you have to prove not only that no higher value than $\frac{8045}{2012}$ can be attained; you also have to prove that $\frac{8045}{2012}$ can be attained. I still don't see any argument to that effect in your answer. I do believe it's true; you just didn't provide an argument for it. – joriki Dec 24 '19 at 09:36
  • Oh, I see where you’re coming from now. Sorry about that. From my past experiences of using inequalities, I’ve just gotten use to finding a constant through rearranging And that is probably my answer. Because I wrote that x+1/x is less than or equal to 8045/2012, I assumed that that was the answer as there was a equal to component to it also. It would be helpful if were able to correct me as I am not very advanced in the use of inequalities. Thanks in advance!!! – JC12 Dec 24 '19 at 10:50
  • You can just solve for x, subtract from 2012, and divide by 2010 to get the rest of the terms. It works out – doingmath Dec 24 '19 at 19:09