Ngày tải lên :
05/06/2014, 18:39
... again, we can get the result J Now, I will present another proof of mine based on this theorem y x z y Since a, b, c > 0, abc = 1, there exists x, y , z > such that a = , b = , c = x then our inequality ... = 1, then 1 + + ≥ a + a +1 b + b +1 c + c +1 a = Proof From the given condition a, b, c > 0, abc = , there exist x, y , z > such that b = c = yz x2 zx And y2 xy z2 then, the inequality ... > Then the inequality is proved J Example (V ình Quý) Let a, b, c > 0, abc = Prove that 1 + + ≤ a − a +1 b − b +1 c − c +1 Solution On Mathlinks inequality forum, I posted the following proof:...