09/05/10 21:59:17
>>888
見かけほど難しくない(?)
左側:
a/(pb+qc) + b/(pc+qa) + c/(pa+qb) - (p+q)^2 (a+b+c)^3 /{9(pb+qc)(pc+qa)(pa+qb)}
= {(2p-q)(2q-p)F_1 + (p+q)(2p-q)G + (p+q)(2q-p)H}/{9(pb+qc)(pc+qa)(pa+qb)} ≧ 0,
ここに
F_1 = a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) = s^3 -4st +9u ≧ 0,
G = F_1 + (st-9u+3⊿)/2 = a(a-b)^2 + b(b-c)^2 + c(c-a)^2 ≧ 0,
H = F_1 + (st-9u-3⊿)/2 = a(a-c)^2 + b(b-a)^2 + c(c-b)^2 ≧ 0,
ここに
s = a+b+c, t = ab+bc+ca, u = abc, 基本対称式
⊿ = (a-b)(b-c)(c-a), 差積
中央と右側:
pb+qc = x, pc+qa = y, pa+qb = z, とおく。
a+b+c = (x+y+z)/(p+q),
よって 相加･調和平均より
(x+y+z)^3 /(9xyz) = (x+y+z){F_0 + 3(xy+yz+zx)}/(9xyz) ≧ (1/3)(x+y+z)(1/x + 1/y + 1/z) ≧ 3,
これを (p+q) で割る。ここに
F_0 = (x-y)(x-z) + (y-z)(y-x) + (z-x)(z-y)
= (x^2 + y^2 + z^2) - (xy+yz+zx)
= (1/2){(x-y)^2 + (y-z)^2 + (z-x)^2} ≧ 0,