證明：（1）先證明,函數(shù)g(x)=x㏑x在(1,+∞)上遞增.求導(dǎo)可得g'(x)=1+㏑x.顯然當(dāng)x＞1時(shí),有g(shù)'(x)=1+㏑x>1,故函數(shù)g(x)在(1,+∞)上遞增.因爾有g(shù)(x)-g(x+1)＜0,即x㏑x-(x+1)㏑(x+1)＜0,(x>1).(2)再證明函數(shù)f(x)=[㏑(x+1)]/㏑x.(x＞1)【注：分子是㏑(x+1),分母是㏑x】在（1,+∞)上遞減.求導(dǎo)并整理可得f'(x)=[x㏑x-(x+1)㏑(x+1)]/[x(x+1)㏑²x].由前面結(jié)果可知,在(1,+∞)上,有f'(x)＜0,故在(1,+∞)上,函數(shù)f(x)=[㏑(x+1)]/㏑x遞減.（3）當(dāng)a＞2時(shí),有a＞a-1＞1.由前結(jié)果可知,f(a-1)＞f(a).===>(㏑a)/㏑(a-1)＞[㏑(a+1)]/㏑a.===>㏒(a-1)[a]＞㏒a[a+1].【注：換底.其中,方括號(hào)[]內(nèi)為真數(shù),前者為底數(shù)】