{an}为等差数列，a1>0 ,d>0,n∈N,n>1 求证：lg(an-1)*lg(an+1)<(lgan)^2
热心网友
∵a10 ,d0,n∈N,n1 ∴an＞0∵4ab<(a+b)^2∴lg(an-1)*lg(an+1)<(1/4)[lg(an-1)+lg(an+1)]^2=(1/4)[lg(an)^2-1]^2＜(1/4)[lg(an)^2]^2=(1/4)[2lg(an)]^2=[lg(an)]^2 得证
热心网友
设an=a，公差是d，则a(n-1)=a-d;a(n+1)=a+dlga(n-1)*lga(n+1)=lg(a-d)*lg(a+d)=<{[lg(a-d)+lg(a+d)]/2}^2=[lg(a^2-d^2)/2]^2<[lg(a^2)/2]^2=(2lga/2)^2=(lga)^2=(lgan)^2