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anonymous

  • 5 years ago

Prove that if a and b positive numbers such that a+b=1, then show that (a+1/a)^2+(b+1/b)^2≥(25/2)

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  1. anonymous
    • 5 years ago
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    Before you type anymore, I'm pretty sure he wants to solve \[\left( a+\frac1a\right)^2+\left( b+\frac1b\right)^2 \ge \frac{25}{2}\]

  2. anonymous
    • 5 years ago
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    Which makes sense, because at a=b=0.5, you get 25/2 as expected

  3. anonymous
    • 5 years ago
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    possibly not solve, but at least use the fact that \[a,b \in (0,1)\] (implied by a+b=1, a>0, b>0) to prove a lower bound of 25/2 on that expression?

  4. anonymous
    • 5 years ago
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    i'm going to sleep now :D have fun with that!

  5. anonymous
    • 5 years ago
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    I really wish I was able to say I understand this question, but I really don't. Thanks for the help anyways :)

  6. anonymous
    • 5 years ago
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    I think something must be wrong...

  7. anonymous
    • 5 years ago
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    Clearly \[a+\frac{1}{a} > a+1>1\] and \[b+\frac{1}{b} > b+1>1\] So \[\left(a+\frac{1}{a} \right)^2 + \left(b+\frac{1}{b}\right)^2 > (a+1)^2 + (b+1)^2\] \[=a^2 + 2a + b^2 + 2b + 2 = a^2 + b^2 + 4 = 5-2ab\] However, \[5-2ab < \frac{25}{2}\] should be clear, since \[10 < 25 + 4ab\] is true by hypothesis: 0<a<1, 0<b<1. I'm convinced something is wrong in the question...

  8. anonymous
    • 5 years ago
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    I guess you can't make 5-2ab > 25/2, since suppose it was, then the last inequality above holds, contradicting it...

  9. anonymous
    • 5 years ago
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    it could be that 25/2 is between them?

  10. anonymous
    • 5 years ago
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    I'm pretty sure the inequality is correct, I just can't think of the proof offhand.

  11. anonymous
    • 5 years ago
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    Between what?

  12. anonymous
    • 5 years ago
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    could it be that \[\left(a+\frac{1}{a} \right)^2 + \left(b+\frac{1}{b}\right)^2 > \frac{25}{2} > (a+1)^2 + (b+1)^2\] ?

  13. anonymous
    • 5 years ago
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    Yes, the right inequality is true by inspection.

  14. anonymous
    • 5 years ago
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    so really I've picked too much of a difference between \[a+\frac{1}{a}>a+1\] \[b+\frac{1}{b}>b+1\]

  15. anonymous
    • 5 years ago
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    Yeah, you can't force it down anymore, since the inequality actually achieves it's minimum at a=b=0.5

  16. anonymous
    • 5 years ago
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    hmm yea

  17. anonymous
    • 5 years ago
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    I don't know!

  18. anonymous
    • 5 years ago
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    I'm still trying a few different things... we'll see, haha

  19. anonymous
    • 5 years ago
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    well I can prove it's greater than 13... bizarrely

  20. anonymous
    • 5 years ago
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    will write it tomorrow if i still think it is right...off to sleep now, it is half 3 in the morning lol

  21. anonymous
    • 5 years ago
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    what I am trying is to take these two inequalities: \[a+\frac{1}{a} > a+1\] \[b+\frac{1}{b} > b+1\] and adding them, to get \[\left[ \left(a+\frac{1}{a}\right) + \left(b+\frac{1}{b}\right)\right]^2 - 2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\] \[> (a+1+b+1)^2 - 2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)= 9- 2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\] and then proceed from there...because you've already showed above that at a=b=0.5 we get the equality, so we only need show the >.

  22. anonymous
    • 5 years ago
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    also another thing is that if one of a (or b) is less than 1/2, then the other must be bigger than 1/2, so maybe use that?

  23. anonymous
    • 5 years ago
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    I think Jensen's inequality might be a way to do it, especially given the condition that a+b=1 is very in line with Jensen's. I've got other work to do though, so I will figure it out later. Also, since it's just in two variables and we have another condition, I'm reasonably sure a straight up calculus "find the minimum" solution will work. Very ugly, but it would work.

  24. anonymous
    • 5 years ago
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    Jensen's looks like a good bet, though I had another idea about Young's inequality which is: \[\frac{a^p}{p} + \frac{b^q}{q} \geq ab\] for \[a,b \geq 0, p,q \geq 1\] where \[\frac{1}{p} + \frac{1}{q} = 1\] And we take \[0<a<1, 0<b<1, p=q=2\] With this method I can prove that the expression on the LHS is > 6 but this is clearly less than 12.5 :(

  25. anonymous
    • 5 years ago
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    Thanks, I finally figured it out... haha... great help though... just might help someone else later on :)

  26. anonymous
    • 5 years ago
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    oh! I'd love to know how you did it if you have time...

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