Math Is Fun Forum

integral

Member

Registered: 2010-02-27

Posts: 5

real analysis problem!!

if a+b=1, where a,b∈ R, prove that (a+1/a)² +(b+1/b)² ≥ 25/2....plz help me solving this problem...thank you all.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: real analysis problem!!

You made a slight typo here:

where a,b∈ R[sup]+[/sup]

If one of a and b is negative, the result breaks down. E.g. try a = 1.5, b = −0.5.

To prove the result for positive real a and b, observe that the function

is convex for x ∈ ℝ[sup]+[/sup]. Hence, by Jensen’s inequality,

QED.

Last edited by JaneFairfax (2010-02-27 03:47:11)

integral

Member

Registered: 2010-02-27

Posts: 5

Re: real analysis problem!!

thnx jane

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: real analysis problem!!

You’re welcome. I just realized however that you posted this as a real-analysis problem, so it may be that you have to use calculus here. While Jensen’s inequality gives a quick solution without using calculus, it may not be what your teacher wants. Do check with your teacher on what is required for this problem.

Last edited by JaneFairfax (2010-02-27 03:52:15)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: real analysis problem!!

Hi Jane and integral;

It's overkill because you can just use the substitution a = 1 - b and then maximize a single variable function but the method of Lagrangian multipliers can be used. This is a little rough and needs fleshing out, it's from memory.

We form the vector equations:

We get the simultaneous set of equations to solve:

From equations 1 and 2 we see that a = b. From the third equation we see that a,b = 1 / 2. When a,b = 1/2 it is a minimum. At = a, b = 1/2 the original equation = 12.5 so we are done.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.