Question

a b c 为正实数,求证a/(a+2b+c)+b/(a+b+2c)+c/(2a+b+c)>=3/4

Answer 1

使用柯西不等式

证明:

a/(a+2b+c)+b/(a+b+2c)+c/(2a+b+c)

=a^2/(a^2+2ab+ca)+b^2/(ab+b^2+2bc)+c^2/(2ac+bc+c^2)

>=(a+b+c)^2/[(a^2+b^2+c^2)+3(ab+bc+ca)]

=3(a+b+c)^2/[3(a+b+c)^2+3(ab+bc+ca)]

>=3(a+b+c)^2/[3(a+b+c)^2+(a+b+c)^2]

=3/4.

故原不等式成立