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        riddles >> medium >> Find minimum over all reals
        
(Message started by: Ronno on Apr 8th, 2010, 9:00pm)
Title: Find minimum over all reals
Post by Ronno on Apr 8th, 2010, 9:00pm
Find the minimum value of the expression
f(a,b,c)=(a+b)^4+(b+c)^4+(c+a)^4- (4/7)*(a^4+b^4+c^4)
as a, b, c vary over the real numbers.
Title: Re: Find minimum over all reals
Post by towr on Apr 9th, 2010, 12:51am
Is this medium?
[hide]It seems obvious that if there is a minimum, then a=b=c, because of the symmetry of the formula. And then it's easily seen they must all be 0.[/hide]
Or am I making some mathematical faux pas here?
Title: Re: Find minimum over all reals
Post by Ronno on Apr 11th, 2010, 1:30am
If it's an unique minimum then a=b=c has to hold. Otherwise it is not necessary. For example, if the factor 4/7 was removed, your argument would still produce 0 as the minimum but the expression would not be always non-negative (e.g. at (3,-1,-1) ).
If you still think it's not worthy of medium, you're welcome to move it to easy.
Title: Re: Find minimum over all reals
Post by rmsgrey on Apr 11th, 2010, 9:51am
The symmetric formula just means that if x,y,z is a solution, so are x,z,y; y,x,z; y,z,x; z,x,y; and z,y,x.

An example of a simpler symmetric expression which manages to have multiple minima, most of which are not symmetric would be:

(x2 + y2 - 1)2

Obviously the minima are at x2 + y2=1 - the circle of unit radius centered on the origin.
Title: Re: Find minimum over all reals
Post by pex on Apr 12th, 2010, 12:31am

on 04/11/10 at 09:51:00, rmsgrey wrote:
The symmetric formula just means that if x,y,z is a solution, so are x,z,y; y,x,z; y,z,x; z,x,y; and z,y,x.

An example of a simpler symmetric expression which manages to have multiple minima, most of which are not symmetric would be:

(x2 + y2 - 1)2

Obviously the minima are at x2 + y2=1 - the circle of unit radius centered on the origin.

And then (x2 + y2 - 1)2 + (x + y)2 has only asymmetric minima...

Perhaps (I haven't really thought about it) an interesting question is; what is the maximum value of k for which
(a + b)4 + (b + c)4 + (c + a)4 - k(a4 + b4 + c4)
does not take negative values? We know from Ronno's post that k < 1. [edit] Actually, it shows that k < 48/83, which is only about .007 more than 4/7... [/edit]
Title: Re: Find minimum over all reals
Post by Obob on Apr 12th, 2010, 4:13am
A proof that the minimum of the function occurs at the origin would probably give insight into that question.  I'd suspect that if k > 4/7 then the function is unbounded in the negative direction.
Title: Re: Find minimum over all reals
Post by pex on Apr 13th, 2010, 8:52am
I have found a slightly larger value numerically: the function is unbounded in the negative direction if and only if k > 0.57784613576525140468749613672912.

[hideb]Note that the objective function is homogeneous of degree four; thus, it is sufficient to find the minimum of
(a+b)4 + (b+c)4 + (c+a)4
over the region
a4 + b4 + c4 = 1.

Using the Lagrange multiplier m, we need to solve
(a+b)3 + (c+a)3 = m a3
(a+b)3 + (b+c)3 = m b3
(b+c)3 + (c+a)3 = m c3
a4 + b4 + c4 = 1

Introducing x = b/a and y = c/a, and eliminating m from the first three conditions:
(1+x)3 + (x+y)3 = x3 [ (1+x)3 + (1+y)3 ]
(1+y)3 + (x+y)3 = y3 [ (1+x)3 + (1+y)3 ]

One trivial solution is (x,y) = (1,1), leading to a=b=c=3-1/4 with function value 16. Another trivial solution is (x,y) = (-1,0) (or the other way around), leading to a=2-1/4, b=-2-1/4, c=0, with value 1.

I had to resort to numerical methods to find the other real solutions. There are three of them:
x=1, y=-2.9506378683979899400582321588892
x=-2.9506378683979899400582321588892, y=1
x=y=-0.33890976954855421150997769094074
All of these lead to the same solution (but with a,b,c ordered differently), with
a=0.99351028288114569673074327517161 and
b=c=-0.33671034101536799238728202302250, with function value 0.57784613576525140468749613672912. It is clear that this is the minimum.

Thus, if k = 0.57784613576525140468749613672912, then (a+b)4 + (b+c)4 + (c+a)4 - k (a4 + b4 + c4) is everywhere nonnegative, and it is zero on all multiples of (a,b,c) = ( 0.99351028288114569673074327517161 ; -0.33671034101536799238728202302250 ; -0.33671034101536799238728202302250 ) and their permutations. For reference, we can scale this vector to (1,x,x), where x is the unique real root of
2x5 + 8x4 + 14x3 + 7x2 + 4x + 1 = 0.
(Obtained by setting x=y and dividing out a factor x-1.)[/hideb]
Title: Re: Find minimum over all reals
Post by Ronno on Apr 13th, 2010, 9:25am
That was straightforward enough. :)
But I would expect that there is a simpler, and of course weaker, solution to the original problem using the factor 4/7.
Title: Re: Find minimum over all reals
Post by SWF on Jun 2nd, 2010, 9:06pm
It is always possible to rename a, b, c, such that a*b>=0, because there are 3 values and at least two of them have the same sign (or at least one number is zero).  f(a,b,c) is the sum of 3 terms, all >=0:

f(a,b,c) = (10/7)*( a^2 + b^2 + c^2 -11/5*a*b  + 14/10(a*c + b*c) )^2
+ 72/7*a*b*(a - b)^2
+ ( sqrt(12/35)*(b*c + a*c) +sqrt(588/35)*a*b )^2

f(a,b,c) must therefore be >=0, and the minimum is zero when a=b=c=0.

Coming up with the above was not easy, but expanding to verify the truth of it is.


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