For all real x,y the value of x 2 + 2 x y + 3 y 2 − 6 x − 2 y cannot be less than − k . k is
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Suppose that x 2 + 2 x y + 3 y 2 − 6 x − 2 y = ( x + y − 3 ) 2 + 2 y 2 + 4 y − 9 = − k has a solution. Then ( x + y − 3 ) 2 = − 2 y 2 − 4 y + ( 9 − k ) ≥ 0 hence 2 y 2 + 4 y + ( k − 9 ) = 2 ( y + 1 ) 2 + ( k − 1 1 ) ≤ 0 Thus k is at most 1 1 . Equality can be achieved when y = − 1 , x = 4 . The answer follows.
It is easy to see that x 2 + 2 x y + 3 y 2 − 6 x − 2 y = ( x + y − 3 ) 2 + 2 ( y + 1 ) 2 − 1 1 . Since ( x + y − 3 ) 2 + 2 ( y + 1 ) 2 ≥ 0 , then x 2 + 2 x y + 3 y 2 − 6 x − 2 y = ( x + y − 3 ) 2 + 2 ( y + 1 ) 2 − 1 1 ≥ − 1 1 . Additionally the equality takes place if x + y − 3 = 0 and y + 1 = 0 , that is, when x = 4 , and y = − 1 . Therefore, the answer to the question is 1 1 .
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(Waiting for a non-calculus approach)