Hơ, ko ai giải à ^^ Thế thì tớ ko khách sáo nữa nhé
Không mất tính tổng quát ta giả sử ta giả sử a=max(a,b,c).
Ta có:
${ \left| {\dfrac{{a - b}}{c} + \dfrac{{b - c}}{a} + \dfrac{{c - a}}{b}} \right| = \left| {\dfrac{{\left( {a - b}\right).\left( {b - c} \right).\left( {c - a} \right)}}{{abc}}}\right| = \dfrac{{\left( {a - b} \right).\left( {a - c} \right)}}{a}.\left| {\dfrac{{b - c}}{{bc}}} \right| \cr = \left[ {\left( {a - 1998} \right).\dfrac{{a.1998 - bc}}{{a.1998}} + \dfrac{{\left({1998 - b} \right).\left( {1998 - c} \right)}}{{1998}}} \right].\left| {\dfrac{{b - c}}{{bc}}} \right| \cr \leq\ \dfrac{{\left( {1998 - b} \right).\left( {1998 - c} \right)}}{{1998}}.\left| {\dfrac{{b - c}}{{bc}}} \right| = \left| {\dfrac{{\left( {1998 - b} \right).\left( {1998 - c} \right).\left( {b - c} \right)}}{{1998.bc}}} \right| \cr}$
Ta xét:
$P = \left| {\dfrac{{\left( {1998 - c} \right).\left( {1998 - b} \right).\left( {b - c} \right)}}{{1998.bc}}} \right|$
Nếu b>c:
Ta có:
${ P = \dfrac{{\left( {1998 - c} \right).\left( {1998 - b} \right).\left( {b - c} \right)}}{{1998.bc}} = \left( {\dfrac{{1998 - c}}{{1998.c}}} \right).\dfrac{{\left( {1998 - b} \right).\left( {b - c} \right)}}{b} \cr = \left( {\dfrac{{1998 - c}}{{1998.c}}} \right).\left( {1998 - \left( {b + \dfrac{{1998.c}}{b}} \right) + c} \right) \leq\ \cr \left( {\dfrac{{1998 - c}}{{1998.c}}} \right).\left( {1998 - 2.\sqrt {b.\dfrac{{1998.c}}{b}} + c} \right) = \dfrac{{\left( {1998 - c} \right).\left( {\sqrt {1998} - \sqrt c } \right)^2 }}{{1998.c}} \cr = \left( {\dfrac{1}{c} - \dfrac{1}{{1998}}} \right).\left( {\sqrt {1998} - \sqrt c } \right)^2 \leq\ \left( {\dfrac{1}{{1997}} - \dfrac{1}{{1998}}} \right).\left( {\sqrt {1998} - \sqrt {1997} } \right)^2 \cr = \left( {\dfrac{1}{{\sqrt {1997} }} - \dfrac{1}{{\sqrt {1998} }}} \right)^2 \cr}$
Nếu c>b: Hoàn toàn tương tự.
Đẳng thức xảy ra khi a=1998, c=1197, $ b = \sqrt {1997.1998}$
Edited by Aye-HL, 25-04-2007 - 20:48.