SOLUTION: prove that for x,y,z>=0, x^2+y^2+z^2>=xy+yz+zx

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Question 461593: prove that for x,y,z>=0, x^2+y^2+z^2>=xy+yz+zx
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the vectors (x, y, z) and ( 1%2Fsqrt%283%29, 1%2Fsqrt%283%29,1%2Fsqrt%283%29).
Then direct application of the Cauchy-Schwartz Inequality gives:

%28x%5E2+%2B+y%5E2+%2B+z%5E2%29%2A1+%3E=+%281%2F3%29%28x%2By%2Bz%29%5E2
3x%5E2+%2B+3y%5E2+%2B+3z%5E2+%3E=+x%5E2+%2B+y%5E2+%2B+z%5E2++%2B+2xy+%2B+2xz+%2B+2yz
2x%5E2+%2B2y%5E2+%2B+2z%5E2+%3E=++2xy+%2B+2xz+%2B+2yz
x%5E2+%2By%5E2+%2B+z%5E2+%3E=++xy+%2B+xz+%2B+yz.
we don't even need the non-negativity conditions for x, y, and z.