1/(n+1)+1/(n+9)
=[(n+1)+(n+9)]/(n+1)(n+9)
=2(n+5)/[(n+5)-4][(n+5)+4]
=2(n+5)/[(n+5)^2-16]
>2(n+5)/(n+5)^2(这一步是分母加上16,分子不变,分数减小)
=2/(n+5)
1/(n+4)>1/(n+5),
所以有 1/(1+n)+1/(4+n)+1/(9+n)>3/(n+5),
所以有3/(n+5)>=1/7
n<=16,
不懂可问
1/(n+1)+1/(n+9)
=[(n+1)+(n+9)]/(n+1)(n+9)
=2(n+5)/[(n+5)-4][(n+5)+4]
=2(n+5)/[(n+5)^2-16]
>2(n+5)/(n+5)^2(这一步是分母加上16,分子不变,分数减小)
=2/(n+5)
1/(n+4)>1/(n+5),
所以有 1/(1+n)+1/(4+n)+1/(9+n)>3/(n+5),
所以有3/(n+5)>=1/7
n<=16,
不懂可问