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!

(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!