Find the largest y such that for all
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1 + x 2 1 ≥ y + x y − x
( y + x ) ≥ ( y − x ) ( 1 + x 2 )
y + x ≥ y − x + x 2 y − x 3
2 x ≥ x 2 y − x 3
x y ≤ 2 + x 2
y ≤ x 2 + x
Define a function f ( x ) = x 2 + x
f ′ ( x ) = − 2 x − 2 + 1
− 2 x − 2 + 1 = 0
x = ± 2
f ′ ′ ( x ) = 4 x − 3
f ′ ′ ( 2 ) = ( 2 ) 3 4 > 0
Therefore the minimum value of x when x>0 is 2 .
So y ≤ 2 2 + 2
y = 2 2 = 0 . 2 8 2 . . .
Just wanted to practice my latex haha