设x,y∈R+,a=x+1/y,b=y+1/z,c=z+1/x,求证a+b+c≥6还有一道哈.已知a,b,c∈R+,求证根号下3分之（a^2+b^2+c^2）≥a+b+c/3

Question

设x,y∈R+,a=x+1/y,b=y+1/z,c=z+1/x,求证a+b+c≥6
还有一道哈.
已知a,b,c∈R+,求证根号下3分之（a^2+b^2+c^2）≥a+b+c/3

Answer 1

1,a+b+c=(x+y+z)+(1/x+1/y+1/z)>=3(xyz)^(1/3)+3/(xyz)^(1/3)>=2√{9(xyz)^(1/3)[1/(xyz)^(1/3)]}=6
2,(a-b)^2+(b-c)^2+(c-a)^2=2a^2+2b^2+2c^2-2ab-2bc-2ca>=0
3(a^2+b^2+c^2)>=a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2
(a^2+b^2+c^2)/3>=(a+b+c)^2/9
√[(a^2+b^2+c^2)/3]>=(a+b+c)/3