anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Before you type anymore, I'm pretty sure he wants to solve \[\left( a+\frac1a\right)^2+\left( b+\frac1b\right)^2 \ge \frac{25}{2}\]
anonymous
  • anonymous
Which makes sense, because at a=b=0.5, you get 25/2 as expected
anonymous
  • anonymous
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?

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anonymous
  • anonymous
i'm going to sleep now :D have fun with that!
anonymous
  • anonymous
I really wish I was able to say I understand this question, but I really don't. Thanks for the help anyways :)
anonymous
  • anonymous
I think something must be wrong...
anonymous
  • anonymous
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
anonymous
  • anonymous
I guess you can't make 5-2ab > 25/2, since suppose it was, then the last inequality above holds, contradicting it...
anonymous
  • anonymous
it could be that 25/2 is between them?
anonymous
  • anonymous
I'm pretty sure the inequality is correct, I just can't think of the proof offhand.
anonymous
  • anonymous
Between what?
anonymous
  • anonymous
could it be that \[\left(a+\frac{1}{a} \right)^2 + \left(b+\frac{1}{b}\right)^2 > \frac{25}{2} > (a+1)^2 + (b+1)^2\] ?
anonymous
  • anonymous
Yes, the right inequality is true by inspection.
anonymous
  • anonymous
so really I've picked too much of a difference between \[a+\frac{1}{a}>a+1\] \[b+\frac{1}{b}>b+1\]
anonymous
  • anonymous
Yeah, you can't force it down anymore, since the inequality actually achieves it's minimum at a=b=0.5
anonymous
  • anonymous
hmm yea
anonymous
  • anonymous
I don't know!
anonymous
  • anonymous
I'm still trying a few different things... we'll see, haha
anonymous
  • anonymous
well I can prove it's greater than 13... bizarrely
anonymous
  • anonymous
will write it tomorrow if i still think it is right...off to sleep now, it is half 3 in the morning lol
anonymous
  • anonymous
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 >.
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 6 but this is clearly less than 12.5 :(
anonymous
  • anonymous
Thanks, I finally figured it out... haha... great help though... just might help someone else later on :)
anonymous
  • anonymous
oh! I'd love to know how you did it if you have time...

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