1.设a+(1/b)=1b+(1/c)=1b不等于1,c不等于1,求c+(1/a)2.设abc=1求证[a/(ab+a+1)]+[b/(bc+b+1)]+[c/[ac+c+1)]=13.已知x+2y-z=21,x-y+2z=12求w=x^2+y^2+z^2的最小值

Question

1.设a+(1/b)=1 b+(1/c)=1 b不等于1 ,c不等于1 ,求c+(1/a)
2.设abc=1 求证[a/(ab+a+1)]+[b/(bc+b+1)]+[c/[ac+c+1)]=1
3.已知x+2y-z=21,x-y+2z=12 求w=x^2+y^2+z^2的最小值

Answer 1

1.a+(1/b)=1,即1/b=1-a
b+(1/c)=1,即b=1-1/c
相乘得:1=(1-a)(1-1/c)
-1/c+a/c=a
c+(1/a)=1
2.由abc=1 得:a/(ab+a+1)=a/(ab+a+abc)=1/(b+1+bc )
a/(ab+a+1) + b/(bc+b+1)=1/(b+1+bc) + b/(bc+b+1 )
=(1+b)/(bc+b+1)=(abc+b)/(bc+b+abc )
=(ac+1)/(c+1+ac )
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)=(ac+c+1)/c+1+ac=1
3.x+2y-z=21
x-y+2z=12
即x+2y=z+21
x-y=12-2z
解得:y=z+3,x=15-z
w=x^2+y^2+z^2
=(15-z)^2+(z+3)^2+z^2
=234-24z+3z^2
=3(z-4)^2+186
所以w=x^2+y^2+z^2的最小值186