08/10/27 01:28:31
>>341
B.4049. a,b,c are positive real numbers, such that ab+bc+ca=t. Prove that
a/(a^2 -bc+3t) + b/(b^2 -ca+3t) + c/(c^2 -ab+3t) ≧ 1/(a+b+c),
(略証)
a+b+c =s, abc =u とおく。
(左辺) - (右辺) = 2{us^3 + (s^2 -4t)t^2} / {(6s^2･t^2 + 8t^3 + us^3)s} ････(*)
= 2{u(s^4 -9t^2) + (s^3 -4st +9u)t^2} / {(6s^2･t^2 + 8t^3 + us^3)s^2}
= 2{u(s^2 +3t)F_0 + t^2･F_1} / {(6s^2･t^2 + 8t^3 + us^3)s^2}
≧ 0,
ここで Schur の不等式 F_0 = s^2 -3t ≧0, F_1 = s^3 -4st +9u ≧0 を使った。
(*) a^2 -bc =A, b^2 -ca =B, c^2 -ab =C とおくと
S = A + B + C = s^2 -3t,
T = AB + BC + CA = -t(s^2 -3t),
U = ABC = us^3 - t^3,
(左辺) = a/(A+3t) + b/(B+3t) + c/(C+3t)
= (aBC + AbC + ABc +9st^2)) / (U +3tT +9t^2･S +27t^3)
= 2{us^3 + (s^2 -4t)t^2}/{(6s^2･t^2 + 8t^3 + us^3)s}.