# How to prove this?

1. Feb 19, 2012

### topengonzo

1. The problem statement, all variables and given/known data
How to prove (a+1/a)^2 + (b+1/b)^2 >= 25/2 given that a+b=1 and a,b positive

2. Relevant equations

3. The attempt at a solution
I tried replacing b with 1-a but I get 6th degree a which I dont 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?

2. Feb 19, 2012

### tiny-tim

hi topengonzo!

have you tried differentiating?

3. Feb 19, 2012

### Dick

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.

4. Feb 19, 2012

### topengonzo

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 cant solve :S

5. Feb 19, 2012

### topengonzo

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

6. Feb 19, 2012

### topengonzo

I just have to prove ab + 2/ab -4 >= 25/2 when a+b=1
How to do it without Lagrange

7. Feb 19, 2012

### Dick

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.

8. Feb 20, 2012

### checkitagain

If you multiply through by the lcd, it will become

$$a^3(2a - 1)(1 - a)^3 - (1 - a)^3 + a^3 = 0$$

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.