數學知識交流---求證

a/(b+c) + b/(a+c) + c/(a+b)
= (a+b+c)/(b+c) + (a+b+c)/(a+c) + (a+b+c)/(a+b) - 3
= (1/2) [ 2(a+b+c)/(b+c) + 2(a+b+c)/(a+c) + 2(a+b+c)/(a+b) ] - 3
= (1/2) [ (a+b + b+c + a+c) / (b+c)
+ (a+b + b+c + a+c) / (a+c)
+ (a+b + b+c + a+c) / (a+b) ] - 3
= (1/2) [ 1 + (a+b)/(b+c) + (a+c)/(b+c)
+ 1 + (a+b)/(a+c) + (b+c)/(a+c)
+ 1 + (b+c)/(a+b) + (a+c)/(a+b) ] - 3
= (1/2) [ 3 + (a+b)/(b+c) + (b+c)/(a+b) + (a+c)/(b+c) + (b+c)/(a+c) +
(a+b)/(a+c) + (a+c)/(a+b) ] - 3
≥ (1/2) [ 3 + 2 + 2 + 2 ] - 3

= 3/2