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

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)

- chestercat

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- 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

Which makes sense, because at a=b=0.5, you get 25/2 as expected

- 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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## More answers

- anonymous

i'm going to sleep now :D have fun with that!

- anonymous

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

- anonymous

I think something must be wrong...

- 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

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

- anonymous

it could be that 25/2 is between them?

- anonymous

I'm pretty sure the inequality is correct, I just can't think of the proof offhand.

- anonymous

Between what?

- 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

Yes, the right inequality is true by inspection.

- 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

Yeah, you can't force it down anymore, since the inequality actually achieves it's minimum at a=b=0.5

- anonymous

hmm yea

- anonymous

I don't know!

- anonymous

I'm still trying a few different things... we'll see, haha

- anonymous

well I can prove it's greater than 13... bizarrely

- 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

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

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

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

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

Thanks, I finally figured it out... haha... great help though... just might help someone else later on :)

- anonymous

oh! I'd love to know how you did it if you have time...

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