Determine the minimal values of ( x + y 1 ) ( x + y 1 − 2 0 1 8 ) + ( y + x 1 ) ( y + x 1 − 2 0 1 8 ) and ( x + y 1 ) ( x + y 1 + 2 0 1 8 ) + ( y + x 1 ) ( y + x 1 + 2 0 1 8 ) for positive reals x and y . What is the sum of these minimal values?
From BxMO 2018
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Multiplying out and regrouping, the first expression is equal to S 1 = [ x + x − 1 − 1 0 0 9 ] 2 + [ y + y − 1 − 1 0 0 9 ] 2 − 2 × 1 0 0 9 2 + 2 ( y x − x y ) 2 so we see that S 1 ≥ − 2 × 1 0 0 9 2 for all x , y > 0 , with equality when x = y = 2 1 ( 1 0 0 9 + 1 0 0 9 2 − 4 ) or x = y = 2 1 ( 1 0 0 9 − 1 0 0 9 2 − 4 ) . Similarly, the second expression is S 2 = [ x + x − 1 + 1 0 0 9 ] 2 + [ y + y − 1 + 1 0 0 9 ] 2 − 2 × 1 0 0 9 2 + 2 ( y x − x y ) 2 Since x + x − 1 ≥ 2 for all x > 0 , it is clear that S 2 ≥ 2 × 1 0 1 1 2 − 2 × 1 0 0 9 2 for all x , y > 0 , with equality when x = y = 1 . This makes the desired answer 2 ( 1 0 1 1 2 − 2 × 1 0 0 9 2 ) = − 2 0 2 8 0 8 2