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Posers and Puzzles

Posers and Puzzles

  1. 09 Apr '08 11:53
    Prove that for reals a, b, c > 0 that
    (a^2+b^2)^2 >= (a+b+c)(a+b-c)(b+c-a)(c+a-b)
    Where ^2 means squared and >= means greater than or equal to.

    I have solved this, and am happy to share the solution but I'm curious to see how you would go about it. My way is long winded and a bit messy, and I'm sure there is a very elegant AM-GM method or something like that. Awaiting your responses!
  2. 09 Apr '08 22:43 / 1 edit
    Originally posted by Dejection
    Prove that for reals a, b, c > 0 that
    (a^2+b^2)^2 >= (a+b+c)(a+b-c)(b+c-a)(c+a-b)
    Where ^2 means squared and >= means greater than or equal to.

    I have solved this, and am happy to share the solution but I'm curious to see how you would go about it. My way is long winded and a bit messy, and I'm sure there is a very elegant AM-GM method or something like that. Awaiting your responses!
    Just an idea...

    If the sum of two of a,b,c is smaller than the third one, then exactly one of a+b-c, b+c-a, c+a-b is negative, so the RHS is negative and the inequality is true.

    Otherwise, a & b & c are the side lengths of a triangle (possibly of area 0). Write the inequality like this:

    a^2+b^2 >= sqrt((a+b+c)(a+b-c)(b+c-a)(c+a-b))

    the RHS is 4 times the are of the triangle (Heron's formula).

    So, we need to prove that in any triangle, the sum of the areas of the squares on two sides is greater then or equal to 4 times the triangle's area. Maybe someone can find a clever geometric proof of this...
  3. 09 Apr '08 23:23
    Very interesting idea. I had thought of the triangle inequality, but was always looking for an algebraic proof (I am not good at geometry). It would be nice to find a geometric proof, it would be much better than my current one, where i actually expand the whole thing out.
  4. 10 Apr '08 07:52
    Originally posted by David113
    Just an idea...

    If the sum of two of a,b,c is smaller than the third one, then exactly one of a+b-c, b+c-a, c+a-b is negative, so the RHS is negative and the inequality is true.

    Otherwise, a & b & c are the side lengths of a triangle (possibly of area 0). Write the inequality like this:

    a^2+b^2 >= sqrt((a+b+c)(a+b-c)(b+c-a)(c+a-b))

    the RHS is 4 ...[text shortened]... ual to 4 times the triangle's area. Maybe someone can find a clever geometric proof of this...
    not sure if this will validly hold up, but given a constant perimeter (a+b+c) consider the case where triangle abc is equilateral.

    this triangle has the greatest area for its given perimeter, which by trig is 1/2(a)(a*sqrt(3)/2). but then 4 times the area of the triangle is sqrt(3)*a^2

    now the sum of the areas on the squares of two of the sides is MINIMIZED when abc is equilateral. that is, if (a+b) is constant, then (a^2 + b^2) is smallest when a and b are equal. but if a=b then a^2+b^2 = 2*a^2 > sqrt(3)a^2.

    rereading this i think it's full of holes. it is very late here haha
  5. Subscriber deriver69
    Keeps
    10 Apr '08 08:40
    Originally posted by David113
    Just an idea...

    If the sum of two of a,b,c is smaller than the third one, then exactly one of a+b-c, b+c-a, c+a-b is negative, so the RHS is negative and the inequality is true.

    Otherwise, a & b & c are the side lengths of a triangle (possibly of area 0). Write the inequality like this:

    a^2+b^2 >= sqrt((a+b+c)(a+b-c)(b+c-a)(c+a-b))

    the RHS is 4 ...[text shortened]... ual to 4 times the triangle's area. Maybe someone can find a clever geometric proof of this...
    To prove that in any triangle, the sum of the areas of the squares on two sides is greater then or equal to 4 times the triangle's area. Can we get away with :

    a^2+b^2>=2ab sin C (A=.5*ab*sin C)

    The maximum the RHS can be is 2ab (sin C=1)

    So worst case scenario is

    a^2+ b^2 >= 2ab

    a^2-2ab+b^2 >=0

    (a-b)^2 >=0 which is always true as a and b are real
  6. 10 Apr '08 08:46 / 6 edits
    here is what I came up with:

    using the fact that (x + y)(x - y) = x^2 - y^2 for all real x, y looking at the RHS's first two terms take x = a + b and y = c we get

    (1) (a + b + c)(a + b - c) = (a + b)^2 - c^2

    and similarly for the other two terms take x = c and y = a - b to get

    (2) (b + c - a)(c + a - b) = c^2 - (a - b)^2

    hence we have

    (3) RHS = {(a + b)^2 - c^2}{c^2 - (a - b)^2}

    now looking at the second term the crutial step is to realize that for all real a,b >= 0

    (4) {c^2 + (a - b)^2} >= {c^2 - (a - b)^2}

    so we have

    (5) {(a + b)^2 - c^2}{c^2 + (a - b)^2} >= {(a + b)^2 - c^2}{c^2 - (a - b)^2}

    now we expand the expression in brackets on LHS in (5) we get

    (6) { (a + b)^2 - c^2 } = a^2 + b^2 + 2ab - c^2 and
    (7) { c^2 + (a - b)^2 } = a^2 + b^2 - 2ab + c^2

    now take x = a^2 + b^2 and y = 2ab - c^2 and we have

    (8) (a^2 + b^2) + (2ab - c^2) and
    (9) (a^2 + b^2) - (2ab - c^2) so after multipliying we get

    (10) (a^2 + b^2)^2 - (2ab - c^2)^2

    Now, clearly (2ab - c^2)^2 >= 0 hence

    (11) (a^2 + b^2)^2 >= (a^2 + b^2)^2 - (2ab - c^2)^2

    now combining (11) and (5) we get

    (a^2 + b^2)^2 >= (a^2 + b^2)^2 - (2ab - c^2)^2 = { (a + b)^2 - c^2 }{ c^2 + (a - b)^2 } >= { (a + b)^2 - c^2 }{ c^2 - (a - b)^2 } = (a+b+c)(a+b-c)(b+c-a)(c+a-b)

    hence the original result follows
  7. 10 Apr '08 12:17
    I think you're right. I really can't read all that algebra, not when it's set out like this. We want Latex!

    Anyway I had expanded the whole thing, and grouped terms so that it was a quadratic in terms of c. I then proved the discriminant was always < 0, and as the the quadratic was positive and had no roots, it was always greater than zero, which led to the result. Messy.