婵＄偑鍊栭崕鎶剿囬敓锟�
（Ⅱ）要证
c?1a?b?1，即证[2(a?b)?1](c?1)?(a?b?1)(a?b?c?1)， ?a?b?c?12(a?b)?12整理得：a?b?2ab?ac?bc?0，亦即证：(a?b)(a?b?c)?0，因为在三角形中
2a?b?c,?a?b?c?0，所以(a?b)(a?b?c)?0成立，则原不等式成立；
（Ⅲ）由（Ⅱ）得：
111?c?11? ?????a?b?c?1(c?1)(a?b?1)c?1?a?b?c?1a?b?1??t?1a?b?111?a?b?11?t?a?b? ??，令，则??2t?12(a?b)?1a?b?1c?1?2(a?b)?1a?b?1??1t2??t?12t2?3t?1即原不等式成立.
1?a?b?11?11111????， ?，所以??31c?1?2(a?b)?1a?b?1?c?1262?(?2)2tt