一道搞笑三元不等式解答.pdf

1、?t?n?) 1. ?a,b,c?0?K,a + b + c = 1,y f(a,b,c) = X cyc 1 a2 + ab + b2 4 + 2 3 y:?c = min(a,b,c),n: 1 b2 + bc + c2 + 1 c2 + ac + a2 22 2ab + ac + bc + 2c2 yr? 1 b2 + bc + c2 + 1 c2 + ac + a2 22 p2ab + ac + bc + 2c2 (a b)2 = 22 ac + bc + 2c2 + 2ab =?d 2 ? 1 a2+ ac + c2 + 1 b2+ bc + c2 4 ac + bc + 2c2+

2、2ab ?2 ? 1 a2 + ac + c2 1 b2 + bc + c2 ?2 O? (b2+bc+c2)(ac+bc+2c2+2ab)+(a2+ac+c2)(ac+bc+2c2+2ab)4(a2+ac+c2)(b2+bc+c2) = (ab)2(ac+bc+2abc2) (b2+ bc + c2) (a2+ ac + c2) = (a b)(a + b + c)2 ?y 2(2ab + ac + bc c2) 2ab + ac + bc + 2c2 (a + b + c)2 (a2+ ac + c2+ b2 + bc + c2)2 duc = min(a,b,c),k 2(2ab + a

3、c + bc c2) 2ab + ac + bc + 2c2 1 9 p a2+ ac + c2+ p b2+ bc + c2 a + b + c ny.! ? LHS 1 a2 + ab + b2 + 22 2ab + ac + bc + 2c2= 1 p(1 c)2 ab + 22 2ab + c2 + c 5? g(u) = 1 p(1 c)2 u + 22 2u + c2 + c ? g0(u) = 1 2(1 c)2 u3/2 22 (2u + c + c2)3/2 ? 2(1 c)2 u 2u + c + c2 = 2(a2+ ab + b2) (2ab + ac + bc + 2c2) = (a2 ac) + (b2 bc) + (a2+ b2 2c2) 0 1 2 22 23/2 0K g0(u) f1k(a b)2,($ ?E|),= 1 b2 + bc + c2 + 1 c2 + ac + a2 22 p2ab + ac + bc + 2c2 k(a b)2 L? 2(a b)2(a + b + c)2 k(a2+ b2+ 2c2+ ac + bc) 1r,?,N8=A (a b)2(a + b + c)2 ?r ?(a b)2,?a = b = c = 1,?k 9 8,?z$?k = 1= 2