柯西不等式

Question

已知a+b+c=1

求证

(1+1/a^2)(1+1/b^2)(1+1/c^2)>=1000

用其他的方法也行

Answer 1

(a+b+c)(1/a+1/b+1/c)>= (a*1/a+b*1/b+c*1/c)^2=9
=>(1/a+1/b+1/c)>= 9
a+b+c>=3(abc)^(1/3)=>abc<=1/27
(1+1/a^2)(1+1/b^2)(1+1/c^2)=(a^2+1)(b^2+1)(c^2+1)/(a^2b^2^c^2)
=[a^2b^2^c^2+(a^2b^2+b^2c^2+c^2a^2)+(a^2+b^2+c^2)+1]/(a^2b^2^c^2)
>=[a^2b^2^c^2+3(abc)^(4/3)+3(abc)^(2/3)+1]/(a^2b^2^c^2)
=1+3(abc)^(-2/3)+3(abc)^(-42/3)+(abc)^(-2)
>=1+3*(1/27)^(-2/3)+3*(1/27)^(-4/3)+(1/27)^(-2)=1000