(1/2)[f(x1)+f(x2)]=(1/2)[lg(1/x1 -1)+lg(1/x2 -1)]
=(1/2)lg(1/x1 -1)(1/x2 -1)
=(1/2)lg[(1-x1)(1-x2)/x1x2]
(1/2)[f(x1)+f(x2)]- f[(x1+x2)/2]
=(1/2)lg[(1-x1)(1-x2)/x1x2] -lg[(x1+x2)/2]
=(1/2)lg[(1-x1)(1-x2)/x1x2] -(1/2)lg[(x1+x2)^/4]
=(1/2)lg[4(1-x1)(1-x2)]/[x1x2(x1+x2)^] 。
。。。。。。。
(1)
只需考察真数部分的大小即可判断正负
4(1-x1)(1-x2)]/[x1x2(x1+x2)^]
因为为,x在(0,1/2)
===>(1-x1)(1-x2)>x1x2
显然4>(x1+x2)^
所以,真数大于1
===>(1)式大于0
===>(1/2)[f(x1)+f(x2)] > f[(x1+x2)/2]。
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