Complete them!
If and are real numbers, find the minimum possible value of the expression above.
The answer is 4.
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1 solution
( x 2 + 1 0 x + 2 7 ) ( y 2 − 6 y + 1 1 ) = ( ( x + 5 ) 2 + 2 ) ( ( y − 3 ) 2 + 2 )
We can see the minimum from this is 4 and occurs when x = − 5 and y = 3
The minimum of the whole expression is the product of the minimum of the two quadratics in this case since they're positive for all values of x and y .