222a,0?y?a,0?z?a. 333121222222证明 由x?y?(a?x?y)?a有x?(y?a)x?(y?ay?a)?0. 其判别
24122222式??(y?a)?4(y?ay?a)?0(因x?R). 从而, 3y?2ay?0即0?y?a.
3422同理可证0?x?a,0?z?a.
330?x?3．设a,b,c表示一个三角形三边的长, 求证:
a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?3abc.
证明不失一般性, 设a?b?c, 令a?c?m,b?c?n, 则m?n?0. 有
3abc?a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?a(a?b)(a?c)?b(b?c)(b?a)?c(c?a)(c?b)?(c?m)(m?n)m?(c?n)n(n?m)?cmn?(m?n)[c(m?n)?(m2?n2)]?cmn?0.
?a2(b?c?a)?b2(c?a?b)?c2(a?b?c)?3abc.
4．设x,y?R, 且x?y?1.求证: x2?2xy?y2?证明 设x?y??, 则由题设可知,
222222.
??1, 并可设x??cos?,x??sin?.于是
x2?2xy?y2??2(cos2??2cos?sin??sin2?)??2(cos2??sin2?)??22sin(??).
4?x2?2xy?y2?2.
5．已知a?1,b?1, 求证
?a?b?1.
1?ab证明 欲证
a?ba?b2?1成立, 只需()?1, 即证(a?b)2?(1?ab)2.
1?ab1?ab222222则只需(1?ab)?(a?b)?0, 也就是1?ab?a?b?0, 即证
(1?a2)(1?b2)?0. 而a?1,b?1, 所以(1?a2)(1?b2)?0成立. 命题得证.
6．若
?ai?1(ai?0), 求?(ai?i?1i?1nn11)?(n?)n. ain证明 a1?11111?a1?2?2?...?2?(n2?1)n2?12n2n2?1, a1na1na1na1na1n2项a2?11111?a2?2?2?...?2?(n2?1)n2?12n2n2?1, …… …… a2na2na2na2na2n2项an?11111?an?2?2?...?2?(n2?1)n2?12n2n2?1. annannannannann2项以上诸式, 当且仅当ai?1(i?1,2,...,n)是等号成立. 诸式两端相乘得n(a1?1111)(a2?)...(an?)?(n2?1)nnn2?12n3.
n2?1a1a2ann(a1a2...an)nn11n2?1n3?n)?n. 由已知?ai?1可得a1a2...an?,(naa...ai?112nn11n11112nn2?1(a?)?(n?). 即(a1?)(a2?)...(an?)?(n?1)n,?i2n3n?n3ana1a2annni?1i等号当且仅当a1?a2?...?an?851时成立. n27．证明: 函数f(x)?x?x?x?x?1?0. 证明 (1) 当x?(??,0)时, 显然f(x)?0;
(2) 当x?(0,1)时, f(x)?x?x(1?x)?(1?x)?0; (3) x?[1,??)时, f(x)?x(x?1)?x(x?1)?1?0. 综合(1), (2), (3)可知, 可知f(x)恒正. 8．证明 若ai?1(i?1,2,...,n),则2证明 用数学归纳法证明如下: 当n?1时, 命题显然成立;
假设命题对n成立, 我们来证明它对n?1也成立, 注意到ai?1(i?1,2,...,n).
n?182353(a1a2...an?1)?(1?a1)(1?a2)...(1?an).
?(1?a)?(1?aii?1n?1n?1)?2n?1(?ai?1)?2i?1nn?1(?ai??ai?an?1?1)i?1i