Solution 1 by turcas c If somebody had the patience to make the common denominator, he would had a sur prise: a 2 b(b − c) a + b + b 2 c(c − a) b + c + c 2 a(a − b) c + a = a 3 b 3 − a 3 b 2 c + a 3 c 3 − a 2 bc3 + b 3 c 3 − ab3 c 2 (a + b)(b + c)(c + a) . We have to prove that: a 3 b 3 − a 3 b 2 c + a 3 c 3 − a 2 bc3 + b 3 c 3 − ab3 c 2 ≥ 0 ⇔ a 3 b 3 + b 3 c 3 + c 3 a 3Solution 1 by turcas c If somebody had the patience to make the common denominator, he would had a sur prise: a 2 b(b − c) a + b + b 2 c(c − a) b + c + c 2 a(a − b) c + a = a 3 b 3 − a 3 b 2 c + a 3 c 3 − a 2 bc3 + b 3 c 3 − ab3 c 2 (a + b)(b + c)(c + a) . We have to prove that: a 3 b 3 − a 3 b 2 c + a 3 c 3 − a 2 bc3 + b 3 c 3 − ab3 c 2 ≥ 0 ⇔ a 3 b 3 + b 3 c 3 + c 3 a 3