Prove that for any triangle with sides a, b, and c, the follow inequalities hold: 3/2 <= a/(b+c) + b/(a+c) + c/(a+b) < 2

PART 1

I worked this one out backwards so it's nothing interesting.

By the triangle inequality theorem, the sum of any two sides is greater than the third.

b+c>a     -->     a^2(b+c)>a^3
a+c>b     -->     b^2(a+c)>b^3
a+b>c     -->     c^2(a+b)>c^3

Along with 2abc>abc. We have

a^2(b+c)+b^2(a+c)+c^2(a+b)+2abc>a^3+b^3+c^3+abc
a^2(b+c)+ab(b+c)+bc(b+c)+ac(b+c)>a^3+b^3+c^3+abc
(b+c)(a^2+ab+bc+ac)>a^3+b^3+c^3+abc
(b+c)(a+b)(a+c)>a^3+b^3+c^3+abc

Adding (a+b)(a+c)(b+c)=a^2(b+c)+b^2(a+c)+c^2(a+b)+2abc to both sides and grouping:

2(a+b)(a+c)(b+c)>a^3+b^3+c^3+3abc+a^2(b+c)+b^2(a+c)+c^2(a+b)

2(a+b)(a+c)(b+c)>(a^3+a^2b+a^2c+abc)+(b^3+ab^2+b^2c+abc)+(c^3+ac^2+bc^2+abc)

2(a+b)(a+c)(b+c)>a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c)

2 > a(b+c) + b/(a+c) + c/(a+b)

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PART 2

This one I used some calculus to solve. But I won't go into details on the math. It's not too hard or long but will be a pain to type out.

Define f(a,b,c)=a/(b+c) + b/(a+c) + c/(a+b)

To find an extrema, the partial derivatives must be 0. After a little manipulation, you will see that the only extremum is a=b=c so that f=3/2. You can easily find another point nearby to show it is a minimum.

To show that it's a global minimum instead of a local minimum, we have to consider the boundaries. Our restraint is that a,b,c>0. If you consider the a=0 (or b=0 or c=0), you will get f=b/c+c/b=(b^2+c^2)/bc>=2. That can be shown with (b-c)^2>=0 --> b^2+c^2>=2bc. Of course, this is greater than 3/2.

To consider the unbounded a,b,c --> infinity. Consider the boundaries to be an arbitrary cube with side d. So that a,b,c<d. If you are to look at the boundaries a=d (or b=d or c=d), you will arrive at a minimum for b=c. It's the same problem but has been reduced to 2D. Doing it again reduces it to 1D and then you can see that it's an absolute minimum inside a cube of arbitrary length d.

a=b=c is a minimum regardless of the size of the cube. Therefore, the results aren't change if d approaches infinity. Therefore, f=3/2 is an absolute minimum.

Posted by np_rt on 2004-08-13 15:34:15