Question

设a,b,c是实数,且a^2+b^2=1,c^2+d^2=1,则abcd的最小值等于多少?

Answer 1

设a=sinx,b=cosx,c=siny,d=cosy,则abcd=sinxcosxsinycosy=(1/4)sin2xsin2y
sin2xsin2y最小值为-1，则abcd最小值为-1/4

Answer 2

a^2+b^2=1
a^2 = 1-b^2

p1= a^2b^2
p1 =(1-b^2)b^2
p1' = 2b - 4b^3 =0
b(2-b^2) = 0
b = 0 , √2/2 or -√2/2
p1''=2 - 12b^2
p1''(√2/2) = p''(-√2/2) > 0 (min)

min p1 at b = √2/2
=> min ab at b = √2/2

similarly
p2= c^2d^2

=> min cd at d =√2/2

min abcd = (√2/2)(√2/2)(√2/2)(√2/2)
= 1/16 #