My proving based on the senior's enlightenment:
1/a=(1/b+1/c)/(1-1/b*c)>=1/b+1/c
then we assume that a<b && a<c because if either of b and c is smaller than a or equal to a , the inequation above will not set
let b=a+i and c=a+j   (i,j>=1)
then a=(bc-1)/(b+c)
     a=((a+i)(a+j)-1)/(2a+i+j)
     2a^2+(i+j)a = a^2+(i+j)a+ij-1
     a^2+1=ij
then get the function we need :
		b+c=2a+i+j=2a+i+(a^2+1)/i   
theoretically when i=sqrt(a^2+1) 
		(b+c) gets minimum value
but actually i is an integer and sqrt(a^2+1) cannot get a value of integer , but the actual i is very close to sqrt(a^2+1) as long as it makes j also an integer value
according to the function's property above that when i is less than sqrt(a^2+1), it is a decreasing function,which means if i decreases the function value increases
let i from a to 1
when (a^2+1)%i ==0 ,which means we have got the value of i for the actual minimum function value
just get the value of i and j=(a^2+1)/i
then get the value of b+c=2a+i+(a^2+1)/i which is supposed to be the minimum value