(1)∵平方平均≥算术平均,∴根号【【(a+1/a)^2+(b+1/b)^2+(c+1/c)^2】/3】≥【a+1/a+b+1/b+c+1/c】/3=【1+1/a+1/b+1/c】/3化简得(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≥【1+1/a+1/b+1/c】*【1+1/a+1/b+1/c】/3当且仅当a+1/a=b+1/b=c+1/c时等号成立.

　　由算术平均≥调和平均可得【1/a+1/b+1/c】/3≥3/【a+b+c】=3化简得1/a+1/b+1/c≥9当且仅当1/a=1/b=1/c,即a=b=c时等号成立.

　　所以(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≥【1+9】*【1+9】/3=100/3当且仅当a+1/a=b+1/b=c+1/c且a=b=c时等号成立.解得a=b=c=1/3

　　(2)∵平方平均≥算术平均,所以【根号（a+1/2）+根号(b+1/3)+根号（c+1/4）】/3≤根号【【a+1/2+b+1/3+c+1/4】/3】=根号【25/36】=5/6

　　整理得根号（a+1/2）+根号(b+1/3)+根号（c+1/4）≤5/2当且仅当a+1/2=b+1/3=c+1/4时等号成立.所以由题目条件易得a=7/36b=13/36c=16/36=4/9