HMMT 二月 2010 · CALC 赛 · 第 7 题

英文解析

7. [ 6 ] Let a , a , and a be nonzero complex numbers with non-negative real and imaginary parts. Find
   1 2 3
   the minimum possible value of
   | a + a + a |
   1 2 3
   √ .
   | a a a |3
   1 2 3
   √ √
   iθ π3
   Answer: 3 2 Write a in its polar form re where 0 ≤ θ ≤ . Suppose a , a and r are fixed so
   1 2 3
   iφ2
   that the denominator is constant. Write a + a as se . Since a and a have non-negative real and
   2 3 2 3
   imaginary parts, the angle φ lies between 0 and . Consider the functionπ
   2 iθ iφ 2 2 22
   f ( θ ) = | a + a + a | = | re + se | = r + 2 rs cos( θ − φ ) + s .
   1 2 3
   π π
   ′′ ′′
   Its second derivative is f ( θ ) = − 2 rs (cos( θ − φ ))). Since − ≤ ( θ − φ ) ≤ , we know that f ( θ ) < 0
   2 2
   and f is concave. Therefore, to minimize f , the angle θ must be either 0 or . Similarly, each of a , aπ
   1 2
   and a must be either purely real or purely imaginary to minimize f and the original fraction.2
   By the AM-GM inequality, if a , a and a are all real or all imaginary, then the minimum value of 3
   1 2 3
   the fraction is 3. Now suppose only two of the a ’s, say, a and a are real. Since the fraction isi 1 2
   homogenous, we may fix a + a - let the sum be 2. The term a a in the denominator acheives its
   1 2 1 2
   maximum only when a and a are equal, i.e. when a = a = 1. Then, if a = ki for some real
   1 2 1 2 3
   number k , then the expression equals
   √
   k + 42
   √ .
   k 3
   √ √
   Squaring and taking the derivative, we find that the minimum value of the fraction is 3 2, attained 3
   √
   when k = 2. With similar reasoning, the case where only one of the a ’s is real yields the sameiminimum value.
   ∞ ∞
   ∑ ∑ 1