| 1 | |
a3 + ab2 + ac2 > a2b + abc + a2c | |
| 3 |
| 1 | |
a3 + ab2 + ac2 − a2b − 1 − a2c = (*) > 0 −− to należy pokazać : | |
| 3 |
| b − a | c − a | |||
(*) > 11 + ab(b − a) + ac(c − a) = 11 + | + | = | ||
| c | b |
| b | a | c | a | |||||
= 11 + | − | + | − | ≥ 11 + 44√b/c * (−a/c) * c/b * (−a/b) = | ||||
| c | c | b | b |
| 1 | ||
a2 > 0, bc > 0 ponieważ a > 0, abc = 1 ⇒ bc = | > 0 | |
| a |