1/ab +1/bc +1/ca + 3/(a+b+c) = c/abc +b/abc+c/abc + 3/(a+b+c) = a+b+c +3/(a+b+c)
= 2(a+b+c)/3 + (a+b+c)/3 + 3/(a+b+c)
áp dụng bđt co sy ta có:
a+b+c >=3.căn bậc 3 (abc) => 2(a+b+c)/3 >= 2
(a+b+c)/3 + 3/(a+b+c) >= 2
=> 1/ab +1/bc +1/ca + 3/(a+b+c) >=4