Question

已知abc为不等正数.求证:1/2a+1/2b+1/2c大于1/(b+c)+1/(a+c)+1/(a+b)

Answer 1

1/4a+1/4b

=(a+b)/4ab

≥(a+b)/(a+b)^2

=1/(a+b)

同理1/4b+1/4c≥1/(b+c)

1/4c+1/4a≥1/(c+a)

由以上三式可得1/2a+1/2b+1/2c≥1/(a+b)+1/(b+c)+1/(c+a)