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...