Prove the following inequality holds for the natural number n  2 :
1 
1

1
 
2 3

1 n
 n
Proof .
We use mathematical induction to prove this inequality.
For the base case n  2 we need to show
Since 2  2 so this implies
1 
1

2
2
1
2

1
2
. Adding 1 to both sides gives
1 
1
1
2
1
2
(*)
We use the arithmetic – geometric mean inequality which is given by a  b
2
 ab provided a  0, b  0
Applying this to the right hand side of the inequality (*) yields
1 
1
1
2
1 2  1
2 2
 2 1
Hence we have our required inequality 1 
1

2
2 .
Thus the base case n  2 is true.
2
Assume the inequality is true for n  k :
1 
1

1
 
2 3

Required to prove the result for n k 1 :
1 k
 k (†)
1 
1

1
 
2 3

Consider the left hand side of this inequality
1 k

1 
1

1
3
  
1 k

 k
1 k  1
 k  1 k
1
 1
 k  k
1
 1
(**)
Writing the right hand inequality with a common denominator gives

k  k
1
 1

 because k 1 k k k  1
 k  k k
1
1 1 k
1
 1 k  1 k k  1
 k  1 k  1 k  1
 k  1





By the rules of indices





Substituting this inequality k 
1 k  1
 k  1 into (**) gives
1 
1

2
1
3
  
1 k

Hence the inequality is true for n k 1 .
k
1
 1
 k 
1 k  1
 k  1
By mathematical induction we have our proven our given result.
■