Everybody loves to root for a nuisance - 2
Let a , b , c , d be real numbers such that b − d ≥ 1 2 . 5 and all zeroes β 1 , β 2 , β 3 , and β 4 of the polynomial P ( x ) = x 4 + a x 3 + b x 2 + c x + d are real. Find the smallest value the product ( β 1 2 + 1 ) ( β 2 2 + 1 ) ( β 3 2 + 1 ) ( β 4 2 + 1 ) can take.
Also try this .
The answer is 132.25.
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2 solutions
Try a = c = 6 . 5 and b = 1 2 . 5 , d = 0 for real zeros 0 , 1 , 1 . 6 4 3 8 9 , 4 . 1 0 6 1 1
I did same!!
This problem is more or less the same as 2014 USAMO 1 .
We can factor β k 2 + 1 as ( β k + i ) ( β k − i ) , where i 2 = − 1 . Then, the problem can be rephrased as k = 1 ∏ 4 ( x k − i ) ( x k + i ) . This is equal to P ( i ) P ( − i ) , or ( b − d − 1 ) 2 + ( a − c ) 2 .
Clearly b − d − 1 is minimized when b − d = 1 2 . 5 , so we must minimize ( a − c ) 2 . We set this to 0 , or a = c . Hence the maximum value is 1 1 . 5 2 , or 1 3 2 . 2 5 . Equality is achieved when a = c and b − d = 1 2 . 5 .
By some algebraic manipulation it can be proven that the product is just
( b − d − 1 ) 2 + ( a − c ) 2 . (Hint: fit in x = i , − i )
This is at least ( 1 2 . 5 − 1 ) 2 + 0 = 1 3 2 . 2 5 .
Challenge: find an equality case!