Precalculus, Chapter 9, 9.4, Section 9.4, Problem 27

To prove 1/sqrt(1)+1/sqrt(2)+1/sqrt(3)+1/sqrt(4)+........1/sqrt(n)>sqrt(n)
n>=2
For n=2, 1/sqrt(1)+1/sqrt(2)~~1.707
sqrt(2)~~1.414
:. 1/sqrt(1)+1/sqrt(2)>sqrt(2)
Let's assume that for k> 2 , 1/sqrt(1)+1/sqrt(2)+1/sqrt(3)+.....+1/sqrt(k)>sqrt(k)
So, 1/sqrt(1)+1/sqrt(2)+1/sqrt(3)+....+1/sqrt(k)+1/sqrt(k+1)>1/sqrt(k+1)
or let's show that sqrt(k)+1/sqrt(k+1)>sqrt(k+1) ,k>2
Multiply the above inequality by sqrt(k+1)
sqrt(k)sqrt(k+1)+1>(k+1)
Rewriting the above inequality as below; shows that the above is true ,
sqrt(k)sqrt(k+1)+1>sqrt(k)sqrt(k)+1
:. 1/sqrt(1)+1/sqrt(2)+1/sqrt(3)+......+1/sqrt(k)+1/sqrt(k+1)>sqrt(k+1)
By the extended mathematical induction, the inequality is valid for all n,n>=2