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Problem 6 (due on Monday, November 18).
For real numbers a,b,c consider the system of equations x2+2yz=a, y2+2xz=b, z2+2xy=c. Prove that this system has at most one solution in real numbers x,y,z such that x≥y≥z and x+y+z≥0. Prove that such a solution exists if and only if a+b+c≥0 and b=min(a,b,c). Here min(a,b,c) denotes the smallest number among a,b,c.
No solution were submitted.