Advanced book on Mathematics Olympiad

2.2 Polynomials 49

Next, a problem from the short list of the 2005 Ibero-American Mathematical
Olympiad.

Example.Find the largest real numberkwith the property that for all fourth-degree
polynomialsP(x)=x^4 +ax^3 +bx^2 +cx+dwhose zeros are all real and positive,
one has

(b−a−c)^2 ≥kd,

and determine when equality holds.

Solution.Letr 1 ,r 2 ,r 3 ,r 4 be the zeros ofP(x). Viète’s relations read

a=−(r 1 +r 2 +r 3 +r 4 ),

b=r 1 r 2 +r 1 r 3 +r 1 r 4 +r 2 r 3 +r 2 r 4 +r 3 r 4 ,

c=−(r 1 r 2 r 3 +r 1 r 2 r 4 +r 1 r 3 r 4 +r 2 r 3 r 4 ),

d=r 1 r 2 r 3 r 4.

From here we obtain

b−a−c=(r 1 r 2 +r 1 r 3 +r 1 r 4 +r 2 r 3 +r 2 r 4 +r 3 r 4 )+(r 1 +r 2 +r 3 +r 4 )

+(r 1 r 2 r 3 +r 1 r 2 r 4 +r 1 r 3 r 4 +r 2 r 3 r 4 ).

By the AM–GM inequality this is greater than or equal to

1414

√

(r 1 r 2 r 3 r 4 )^7 = 14

√

d.

Since equality can hold in the AM–GM inequality, we conclude thatk =196 is the
answer to the problem. Moreover, equality holds exactly whenr 1 =r 2 =r 3 =r 4 =1,
that is, whenP(x)=x^4 − 4 x^3 + 6 x^2 − 4 x+1. 

And now a challenging problem from A. Krechmar’sProblem Book in Algebra(Mir
Publishers, 1974).

Example.Prove that

3

√

cos

2 π

7

+^3

√

cos

4 π

7

+^3

√

cos

8 π

7

=^3

√

1

2

( 5 − 33

√

7 ).

Solution.We would like to find a polynomials whose zeros are the three terms on the left.
Let us simplify the problem and forget the cube roots for a moment. In this case we have
to find a polynomial whose zeros are cos^27 π, cos^47 π, cos^87 π. The seventh roots of unity
come in handy. Except forx=1, which we ignore, these are also roots of the equation
x^6 +x^5 +x^4 +x^3 +x^2 +x+ 1 =0, and are cos^2 kπ 7 +isin^2 kπ 7 ,k= 1 , 2 ,...,6. We