Problem: n ≥ 2 n ≥ 2 a 1 , a 2 , … , a n a 1 , a 2 , … , a n
( a 1 + a 2 + ⋯ + a n ) ( 1 a 1 + 1 a 2 + ⋯ + 1 a n ) ≤ ( n + 1 2 ) 2 . ( a 1 + a 2 + ⋯ + a n ) ( a 1 1 + a 2 1 + ⋯ + a n 1 ) ≤ ( n + 2 1 ) 2 .
Prove that max ( a 1 , a 2 , … , a n ) ≤ 4 min ( a 1 , a 2 , … , a n ) max ( a 1 , a 2 , … , a n ) ≤ 4 min ( a 1 , a 2 , … , a n )
Solution:
Let m = min ( a 1 , a 2 , … , a n ) m = min ( a 1 , a 2 , … , a n ) M = max ( a 1 , a 2 , … , a n ) M = max ( a 1 , a 2 , … , a n ) a 1 = m a 1 = m a n = M a n = M m = min ( a 1 , a 2 , … , a n ) m = min ( a 1 , a 2 , … , a n ) M = max ( a 1 , a 2 , … , a n ) M = max ( a 1 , a 2 , … , a n ) m = a 1 ≤ a 2 ≤ ⋯ ≤ a n = M m = a 1 ≤ a 2 ≤ ⋯ ≤ a n = M n = 2 n = 2
If n = 2 n = 2
( m + M ) ( 1 m + 1 M ) ≤ 25 4 ( m + M ) ( m 1 + M 1 ) ≤ 4 2 5
It follows that
4 ( m + M ) 2 ≤ 25 M m or ( 4 M − m ) ( M − 4 m ) ≤ 0 (1) 4 ( m + M ) 2 ≤ 2 5 M m or ( 4 M − m ) ( M − 4 m ) ≤ 0 ( 1 )
Because 4 M − m > 0 4 M − m > 0 M − 4 m ≤ 0 M − 4 m ≤ 0 M ≤ 4 m M ≤ 4 m
We may assume from now that n ≥ 3 n ≥ 3
. The Cauchy-Schwarz Inequality gives
( n + 1 2 ) 2 ≥ ( a 1 + a 2 + ⋯ + a n ) ( 1 a 1 + 1 a 2 + ⋯ + 1 a n ) = ( m + a 2 + ⋯ + a n − 1 + M ) ( 1 M + 1 a 2 + ⋯ + 1 a n − 1 + 1 m ) ≥ ( m M + 1 + ⋯ + 1 ⏟ n − 2 + M m ) 2 ( n + 2 1 ) 2 ≥ ( a 1 + a 2 + ⋯ + a n ) ( a 1 1 + a 2 1 + ⋯ + a n 1 ) = ( m + a 2 + ⋯ + a n − 1 + M ) ( M 1 + a 2 1 + ⋯ + a n − 1 1 + m 1 ) ≥ ( M m + n − 2 1 + ⋯ + 1 + m M ) 2
Hence
n + 1 2 ≥ m M + n − 2 + M m or m M + M m ≤ 5 2 (2) n + 2 1 ≥ M m + n − 2 + m M or M m + m M ≤ 2 5 ( 2 )
It follows that
2 ( m + M ) ≤ 5 M m 2 ( m + M ) ≤ 5 M m
which is (1), completing our proof.
Solution 2. Consider the quadratic polynomial (in x x
p ( x ) = 1 2 [ ( a 1 x + 1 a n ) 2 + ( a n x + 1 a 1 ) 2 + ∑ i = 2 n − 1 ( a i x + 1 a i ) 2 + ( 5 − 2 m M − 2 M m ) x ] = ( 1 2 ∑ i = 1 n a i ) x 2 + 2 n + 1 2 ⋅ x + ( 1 2 ∑ i = 1 n 1 a i ) p ( x ) = 2 1 [ ( a 1 x + a n 1 ) 2 + ( a n x + a 1 1 ) 2 + i = 2 ∑ n − 1 ( a i x + a i 1 ) 2 + ( 5 − 2 M m − 2 m M ) x ] = ( 2 1 i = 1 ∑ n a i ) x 2 + 2 2 n + 1 ⋅ x + ( 2 1 i = 1 ∑ n a i 1 )
Its discriminant is equal to
Δ = ( n + 1 2 ) 2 − ( ∑ i = 1 n a i ) ( ∑ i = 1 n 1 a i ) Δ = ( n + 2 1 ) 2 − ( i = 1 ∑ n a i ) ( i = 1 ∑ n a i 1 )
which, by the given condition is nonnegative. Thus p ( x ) p ( x ) r r
0 = 2 p ( r ) ≥ ( 5 − 2 m M − 2 M m ) r 0 = 2 p ( r ) ≥ ( 5 − 2 M m − 2 m M ) r
Because all of the coefficients of p p r < 0 r < 0
Solution 3. We set a = a 2 + ⋯ + a n − 1 n − 2 a = n − 2 a 2 + ⋯ + a n − 1 m ≤ a 2 ≤ a ≤ a n − 1 ≤ M m ≤ a 2 ≤ a ≤ a n − 1 ≤ M a 2 + ⋯ + a n − 1 = a 2 + ⋯ + a n − 1 = ( n − 2 ) a ( n − 2 ) a
1 a 2 + ⋯ + 1 a n − 1 ≥ ( n − 2 ) 2 a 2 + ⋯ + a n − 1 = n − 2 a a 2 1 + ⋯ + a n − 1 1 ≥ a 2 + ⋯ + a n − 1 ( n − 2 ) 2 = a n − 2
If follows that
( n + 1 2 ) 2 ≥ ( a 1 + a 2 + ⋯ + a n ) ( 1 a 1 + 1 a 2 + ⋯ + 1 a n ) ≥ ( m + ( n − 2 ) a + M ) ( 1 m + n − 2 a + 1 M ) = ( m + M ) ( 1 m + 1 M ) + ( n − 2 ) 2 + ( n − 2 ) ( m + M ) a + ( n − 2 ) a ( 1 m + 1 M ) = ( m + M ) 2 m M + ( n − 2 ) 2 + ( n − 2 ) ( m + M ) m M ⋅ ( m M a + a ) ( n + 2 1 ) 2 ≥ ( a 1 + a 2 + ⋯ + a n ) ( a 1 1 + a 2 1 + ⋯ + a n 1 ) ≥ ( m + ( n − 2 ) a + M ) ( m 1 + a n − 2 + M 1 ) = ( m + M ) ( m 1 + M 1 ) + ( n − 2 ) 2 + a ( n − 2 ) ( m + M ) + ( n − 2 ) a ( m 1 + M 1 ) = m M ( m + M ) 2 + ( n − 2 ) 2 + m M ( n − 2 ) ( m + M ) ⋅ ( a m M + a )
By the AM-GM Inequality, we have m M a + a ≥ 2 m M a m M + a ≥ 2 m M m ≤ a = m M ≤ m ≤ a = m M ≤ M M
( n + 1 2 ) 2 ≥ ( m + M ) 2 m M + ( n − 2 ) 2 + 2 ( n − 2 ) ( m + M ) m M ( n + 2 1 ) 2 ≥ m M ( m + M ) 2 + ( n − 2 ) 2 + m M 2 ( n − 2 ) ( m + M )
Setting t = m + M m M t = m M m + M
( n + 1 2 ) 2 ≥ t 2 + ( n − 2 ) 2 + 2 ( n − 2 ) t = ( t + n − 2 ) 2 ( n + 2 1 ) 2 ≥ t 2 + ( n − 2 ) 2 + 2 ( n − 2 ) t = ( t + n − 2 ) 2
from which it follows that
n + 1 2 ≥ t + n − 2 n + 2 1 ≥ t + n − 2
Hence t ≤ 5 / 2 t ≤ 5 / 2
This problem was suggested by Titu Andreescu. The second solution was contributed by Adam Hesterberg, and the third by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions