Problem:
Let a,b,c,d be real numbers such that b−d≥5 and all zeros x1​,x2​,x3​, and x4​ of the polynomial P(x)=x4+ax3+bx2+cx+d are real. Find the smallest value the product (x12​+1)(x22​+1)(x32​+1)(x42​+1) can take.
Solution:
Using Vieta's identities we have:
x1​x2​+x1​x3​+x1​x4​+x2​x3​+x2​x4​+x3​x4​−x1​x2​x3​x4​≥5
and so
x1​(x2​+x3​+x4​−x2​x3​x4​)+1(x2​x3​+x2​x4​+x3​x4​−1)≥4
It follows that
42≤[x1​(x2​+x3​+x4​−x2​x3​x4​)+1(x2​x3​+x2​x4​+x3​x4​−1)]2
and by the Cauchy-Schwarz Inequality,
42≤(x12​+1)[(x2​+x3​+x4​−x2​x3​x4​)2+(x2​x3​+x2​x4​+x3​x4​−1)2]=(x12​+1)(x22​+1)(x32​+1)(x42​+1)​
The equality holds if and only if
x1​(x2​x3​+x2​x4​+x3​x4​−1)=1(x2​+x3​+x4​−x2​x3​x4​)
which is equivalent to
x1​x2​x3​+x1​x2​x4​+x1​x3​x4​+x2​x3​x4​=x1​+x2​+x3​+x4​
that is, a=c. Taking x1​=…=x4​=1 we obtain b−d=5 and that the smallest value of the product in question is 16 .
An alternative, shorter argument runs as follows: we have
(x12​+1)(x22​+1)(x32​+1)(x42​+1)=P(i)P(−i)=((1−b+d)+i(c−a))(1−b+d−i(c−a))=(b−d−1)2+(c−a)2≥16​
with equality if and only if b−d=5 and a=c, both attained if x1​=…=x4​=1.
This problem and solutions were suggested by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions