`1/sqrt(2) + 1/sqrt(3) + 1/sqrt(4) + ... 1/sqrt(n) > sqrt(n), n>= 2` Use mathematical induction to prove the inequality for the indicated values of n.

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`