IMO 1977 LL VIE60

Origin: VIE

Problem

Suppose x0, x1, . . . , xn are integers and x0 > x1 > \cdot \cdot \cdot > xn. Prove that at least one of the numbers |F(x0)|, |F(x1)|, |F(x2)|, . . . , |F(xn)|, where F(x) = xn + a1xn−1 + \cdot \cdot \cdot + an, ai \inR, i = 1, . . . , n, is greater than n! 2n .

Solution

By Lagrange’s interpolation formula we have F(x) = n  j=0 F(xj) $ i̸=j(x −xj) $ i̸=j(xi −xj). Since the leading coefficient in F(x) is 1, it follows that 1 = n  j=0 F(xj) $ i̸=j(xi −xj) . Since

i̸=j (xi −xj)

= j−1

i=0 |xi −xj| n

i=j+1 |xi −xj| \geqj!(n −j)!,

we have 1 \leq n  j=0 |F(xj)| $ i̸=j(xi −xj)

\leq1 n! n  j=0 n j  |F(xj)| \leq2n n! max |F(xj)|. Now the required inequality follows immediately.