证明:可两次运用Cauchy不等式:(抄错题?是下式吧?)(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=1/3*(1^2+1^2+1^2)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=1/3×[1×(a+1/a)+1×(b+1/b)+1×(c+1/c)]^2=1/3×[1+(a+b+c)(1/a+1/b+1/c)]^2>=1/3×[1+(1+1+1)^ 展开
证明:可两次运用Cauchy不等式:(抄错题?是下式吧?)(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=1/3*(1^2+1^2+1^2)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=1/3×[1×(a+1/a)+1×(b+1/b)+1×(c+1/c)]^2=1/3×[1+(a+b+c)(1/a+1/b+1/c)]^2>=1/3×[1+(1+1+1)^2]^2=100/3证毕.补充另证:(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=(a^2+b^2+c^2)+(1/a^2+1/b^2+1/c^2)+6而1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=<3(a^2+b^2+c^2)即a^2+b^2+c^2>=1/3又abc=<[(a+b+c)/3]^3=(1/3)^3=1/27所以1/(abc)>=27故1/a^2+1/b^2+1/c^2>=3(1/abc)^(1/3)=27因此,(a+1/a)^2+(b+1/b)^2+(c+1/c)^2>=1/3+27+6=100/3. 收起
