已知a>0 ,a^2减2ab+c^2=0 ,bc>a^2 , 比较a.b.c大小

(a-c)^2=2ab-2ac>=0
b>=c

bc>a^2得出a<c
于是a<c<=b

a^2+c^2=2ab>0,b>0
a^2-2ac+c^2=2a(b-c)
(a-c)^2=2a(b-c)>=0,b>=c
b^2>bc>a^2
b>a
a^2-2ab+c^2<a^2-2a^2+c^2=c^2-a^2
c^2-a^2>0,c>a
(a-c)^2>0,b>c

b>c>a

a^2-2ab+c^2=0

a^2+c^2=2ab>0

∵a>0,∴b>0

∵bc>a^2>0,∴c>0

因为bc>a^2，所以a处于这2个数的中间,
即b>a>c或c>a>b,

a^2-2ab+c^2=0,
a^2+c^2=2ab>2ac
所以：b>c

所以b>a>c