Proving (a+1/a)^2 + (b+1/b)^2 ≥ 25/2 Given a+b=1

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Homework Help Overview

The discussion revolves around proving the inequality \((a+1/a)^2 + (b+1/b)^2 \geq \frac{25}{2}\) under the condition that \(a + b = 1\) with \(a\) and \(b\) being positive. Participants explore various mathematical approaches to tackle this problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss substituting \(b\) with \(1-a\) and the resulting complexity of the equations. There are mentions of using Lagrange multipliers and differentiation as potential methods. Some express difficulty in solving the equations derived from these methods.

Discussion Status

There is an ongoing exploration of different approaches, including differentiation and critical point analysis. Some participants suggest that critical points may occur where \(a = b\), while others are questioning the existence of additional roots in the equations derived from Lagrange multipliers. No consensus has been reached yet.

Contextual Notes

Participants note the symmetry in the problem and the significance of the case where \(a = b = 0.5\). There are also discussions about the implications of the positivity of \(a\) and \(b\\, and the challenges posed by the equations resulting from the methods attempted.

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Homework Statement


How to prove (a+1/a)^2 + (b+1/b)^2 >= 25/2 given that a+b=1 and a,b positive


Homework Equations





The Attempt at a Solution


I tried replacing b with 1-a but I get 6th degree a which I don't know how to find inequality for.
Since a and b are perfect symmetric in this problem then a=b=0.5 is of interest. I get that its the minimum.
I tried Lagrange multiplier but I get an equation with 3rd degree. By trial and error, I got solution.
Is there any other way of finding solution? By expanding it?
 
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hi topengonzo! :wink:

have you tried differentiating? :smile:
 
Can you show what kind of equations you got from the Lagrange multiplier technique? I think you can argue that a critical point only occurs where a=b, even if you can't explicitly solve the equations.
 
tiny-tim said:
hi topengonzo! :wink:

have you tried differentiating? :smile:

I differentiated it with respect to a and i get the equation

a - 1/(a^3) - (1-a) + 1/(1-a)^3 = 0 which I can't solve :S
 
Dick said:
Can you show what kind of equations you got from the Lagrange multiplier technique? I think you can argue that a critical point only occurs where a=b, even if you can't explicitly solve the equations.

the equations are:

2(a^4) - λ (a^3) -2 = 0
2(b^4) - λ (b^3) -2 = 0
a+b-1=0

I can solve by trial and error to get a=b=0.5 but what if there is another root for these equations
 
I just have to prove ab + 2/ab -4 >= 25/2 when a+b=1
How to do it without Lagrange
 
If (a+1/a)^2 + (b+1/b)^2 has a critical point, then if you change the sign of a or b you still have a critical point. Let's just look for critical points where a>0 and b>0. If you solve for λ you have λ=2*(a^4-1)/a^3 and λ=(b^4-1)/b^3. You want to prove f(x)=2*(x^4-1)/x^3 is a 1-1 function when x>0, so if f(a)=f(b) shows a=b. Take the derivative of f(x) and show it's always positive. That would show it's a 1-1 function.
 
topengonzo said:
i differentiated it with respect to a and i get the equation

a - 1/(a^3) - (1-a) + 1/(1-a)^3 = 0 which i can't solve :s

If you multiply through by the lcd, it will become

[tex]a^3(2a - 1)(1 - a)^3 - (1 - a)^3 + a^3 = 0[/tex]


You may be able to see a solution, or expand it and use the

Rational Root theorem.


Then, in part, you can test values near that value less than and greater
than it to see if it corresponds to a relative minimum.
 

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