It does not matter whether equality or an inequality is there in the equation, it’s the same approach for both. Approach should be such that given statement has to be proved true in the end. Remember to prove your statement correct for at least one value. Let us learn how to prove inequalities with Mathematical Induction through an example: Suppose we have the following inequality d² <= 2 d + 1 for arbitrary values of ‘d’.

To start with the proof we 1st evaluate the expression for d = 1, we get: 1² <= 2 * 1 + 1 or 1 < 3 and its true.

Then we consider d = k for which this statement stands true i.e. k² <= 2 k + 1. Final statement we need to prove is i.e. d² <= 2 d + 1 is true for k + 1 also i.e. (k + 1)² <= 2 (k + 1) + 1.

This can be done as follows:

(k + 1)² <= 2 (k + 1) + 1,

k² + 2 k + 2 <= 2 k + 3,

We know for d = k if k² = 2 k + 1;

2 k + 1 + 2 k + 2 <= 2 k + 3,

We know that 4 k > 2k because k > 1;

4 k + 3 < 2 k + 3.

Thus we have proved that statement d² <= 2 d + 1 is true for d = k + 1.

Hence, we prove that our hypothesis is wrong and our given statement d² <= 2 d + 1 is true.