Question

已知正数a,b,c 满足a+b+c=1,则(1/a+1/b+4/c)的最小值是

Answer 1

16

因a,b,c∈R+, a+b+c=1
所以
1/a+1/b+4/c = (a+b+c)/a + (a+b+c)/b + 4(a+b+c)/c
= 1+b/a+c/a + a/b+1+c/b + 4a/c+4b/c+4
= 6+ (b/a+a/b) + (c/a+4a/c) + (c/b+4b/c)
由基本不等式：a,b∈R+, a+b≥2√(ab), 有
b/a+a/b ≥ 2
c/a+4a/c ≥ 4
c/b+4b/c ≥ 4
当且仅当 a = b = 1/2c 时取最小值
则
1/a+1/b+4/c = 6+ (b/a+a/b) + (c/a+4a/c) + (c/b+4b/c) ≥ 6+2+4+4 = 16
即(1/a+1/b+4/c)的最小值是
(1/a+1/b+4/c)min = 16
当
a = 1/4
b = 1/4
c = 1/2 时取得.