If (a+b)2+(b+c)2+(c+d)2=4(ab+bc+cd) Prove a=b=c=d

⚑ Flag

Expand:a^2+2ab+b^2+b^2+2bc+c^2+c^2+2cd+d^2=4ab+4bc+4cdSubtract 4ab+4bc+4cd from both sides:a^2-2ab+b^2+b^2-2bc+c^2+c^2-2cd+d^2=0Factor:(a-b)^2+(b-c)^2+(c-d)^2=0Since the square of any real number is greater than or equal to 0, none of the three (   )^2 terms can be negative, so they all must be 0.Therefore a=b, b=c, and c=d, and thus a=b=c=d.This is only true for real numbers a, b, c, and d. If a, b, c and d are complex numbers, then you can't determine a single solution.