⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 + 2 x 2 1 + 1 + 2 y 2 1 = 1 + 2 x y 2 x x ( 1 − 2 x ) + y ( 1 − 2 y ) = 3 2 0 0 1 1 9 ⋅ 2 3 9
If x and y satisfy the system of equations above, find the maximum value of x + y .
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The question gives the conditions: 1 + 2 x y > 0 , x ( 1 − 2 x ) ≥ 0 and y ( 1 − 2 y ) ≥ 0 . By solving them, it is obtained that 0 ≤ x ≤ 2 1 and 0 ≤ y ≤ 2 1 .
Below, we will show that for 0 ≤ x , y ≤ 2 1 , 1 + 2 x 2 1 + 1 + 2 y 2 1 ≤ 1 + 2 x y 2 ⇔ 1 + 2 x 2 1 + 1 + 2 y 2 1 + ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) 2 ≤ 1 + 2 x y 4
⇔ ( 1 + 2 x 2 1 + 1 + 2 y 2 1 − 1 + 2 x y 2 ) + ( 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 − 1 + 2 x y 2 ) ≤ 0
⇔ − ( 1 + 2 x y ) ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) 2 ( x − y ) 2 ( 1 − 2 x y ) + ( 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 − 1 + 2 x y 2 ) ≤ 0
We know that ( 1 + 2 x y ) ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) 2 ( x − y ) 2 ( 1 − 2 x y ) ≤ 0 So we have to prove that the second term on the left hand side is less than 0. 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 ≤ 1 + 2 x y 2
⇔ 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 ≤ 4 x 2 y 2 + 4 x y + 1 2 This implies that 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 ≥ 4 x 2 y 2 + 4 x y + 1 , which is obviously true. Hence, our claim is true, with equality holding when x = y . Substituting this into the second equation and solving, we get the maximum values to be x = y = 0 . 4 8 7 5 Hence, the solution is thus found.