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How many ordered real triples ( x , y , z ) satisfy the following equation - 3 x 2 + 3 y 2 + 3 z 2 = 2 x y + 2 y z + 2 z x ?
The answer is 1.
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3 solutions
My solution is nearly the same as @Dheeman Kuaner 's:
( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 = x 2 + y 2 + z 2 − 2 x y − 2 y z − 2 x z = x 2 + y 2 + z 2 − 3 x 2 − 3 y 2 − 3 z 2 = − 2 ( x 2 + y 2 + z 2 )
( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 has to be non-negativ, and − 2 ( x 2 + y 2 + z 2 ) has to be non-positive (since a square number can't be negativ), so the equation will only be true, if ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 = − 2 ( x 2 + y 2 + z 2 ) = x = y = z = 0
I used the Cauchy-Schwarz inequality to determine 2 x y + 2 y z + 2 z x ≤ 2 x 2 + 2 y 2 + 2 z 2 . And if one of x , y , z are nonzero, then 2 x y + 2 y z + 2 z x ≤ 2 x 2 + 2 y 2 + 2 z 2 < 3 x 2 + 3 y 2 + 3 z 2 . And, hence, they can only be equal if x = y = z = 0 .