taking
6abc to the left and reordering we get
(a^2b^2-2bac
+ c^2) + (a^2c^2-2abc + b^2) + (b ^2c^2- 2abc + a^2)
=
(ab-c)^2 + (ac-b)^2 + (bc-a)^2 \ge 0
the
expression can be zero only if ab= c, ac = b , bc = a or abc = 1 or
zero
if
a,b,c all are different then above cannot be zero
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