Question

(a/(b+2c+3d))+(b/(c+2d+3a))+(c/d+2a+3b))+(d/(a+2b+3c))>=2/3 不等式证明

Answer 1

b+2c+3d=t-----b=t-2c-3d
c+2d+3a=z -----c=z-2d-3a
d+2a+3b=m ------d=m-2a-3b
a+2b+3c=n

d+2a+3b=d+2a+3t-6c-9d=2a+3t-6c-8d=m
c+2d+3a=a+2t-4c-3d+3c=a+2t-c-3d=a+2t-z+2d+3a-3d=4a+2t-z-d=6a+2t-z-m+3b=n
6a+3b=z+m+n-2t=c+2d+3a+d+2a+3b+a+2b+3c-2b-4c-6d=6a-3d+3b -----d=0
将d=0代入原式，得
a/(b+2c)+b/(c+3a)+c/（2a+3b)
接下来用不等式公式

Answer 2

柯西不等式