a>1证loga(a+1)>log(a+1)(a+2)

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a>1证loga(a+1)>log(a+1)(a+2)

a>1证loga(a+1)>log(a+1)(a+2)
a>1证loga(a+1)>log(a+1)(a+2)

a>1证loga(a+1)>log(a+1)(a+2)
log(a+1)(a+2)=loga(a+2)/loga(a+1).所以要证的就是loga(a+1)>loga(a+2)/loga(a+1),因为loga(a+1)>0,所以要证的是[loga(a+1)]²>loga(a+2).即a^{[loga(a+1)]²}>a^loga(a+2),即(a+1)^(loga(a+1))>a+2,即(a+1)^(loga(a+1)-1)>(a+1)^0,即loga(a+1)-1>0,即loga(a+1)>1.显然loga(a+1)>1,以上各步骤步步可逆,证毕.