A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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)
anonymous
 4 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)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Before 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
 4 years ago
Best ResponseYou've already chosen the best response.0Which makes sense, because at a=b=0.5, you get 25/2 as expected

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0possibly 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm going to sleep now :D have fun with that!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I really wish I was able to say I understand this question, but I really don't. Thanks for the help anyways :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think something must be wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Clearly \[a+\frac{1}{a} > a+1>1\] and \[b+\frac{1}{b} > b+1>1\] So \[\left(a+\frac{1}{a} \right)^2 + \left(b+\frac{1}{b}\right)^2 > (a+1)^2 + (b+1)^2\] \[=a^2 + 2a + b^2 + 2b + 2 = a^2 + b^2 + 4 = 52ab\] However, \[52ab < \frac{25}{2}\] 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess you can't make 52ab > 25/2, since suppose it was, then the last inequality above holds, contradicting it...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it could be that 25/2 is between them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm pretty sure the inequality is correct, I just can't think of the proof offhand.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, the right inequality is true by inspection.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so really I've picked too much of a difference between \[a+\frac{1}{a}>a+1\] \[b+\frac{1}{b}>b+1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, you can't force it down anymore, since the inequality actually achieves it's minimum at a=b=0.5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm still trying a few different things... we'll see, haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well I can prove it's greater than 13... bizarrely

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0will write it tomorrow if i still think it is right...off to sleep now, it is half 3 in the morning lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what 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
 4 years ago
Best ResponseYou've already chosen the best response.0also 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0Jensen'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 :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, I finally figured it out... haha... great help though... just might help someone else later on :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh! I'd love to know how you did it if you have time...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.