bk=kc-a-3

b=(kc-a-3)/k

a(a- 1)/(b+c)=k

a(a- 1)=bk+ck

bk=(a^2-a-ck)

b=(a^2-a-ck)/k

So (a 2-a-CK)/k = (KC-a-3)/k

a^2-a-ck=kc-a-3

2kc=a^2+3

k=(a^2+3)/2c

(2)k=(a^2+3)/2c & gt； 0

So (a+3)/(c-b) = k >; 0(a+3 & gt； 0)

c-b & gt； 0

c & gtb

(3)k=(a^2+3)/2c=2

a^2+3=4c

A 2-4a+3 = 4c-4a = 4 (c-a) ... (1)

(a-3)(a- 1)=4(c-a)

a(a- 1)= 2 b+ 2c & lt； 4c =a^2+3

a^2-a<； a^2-3

A>3 So (a-3) (a- 1) = 4 (c-a) is represented by formula (a)>0.

So c> is a c>b.

So c is the maximum.