Problem:

1. Prove the inequality —- for all natural numbers greater than or equal to 1.

Proof 1:

First consider the following: . Let us prove this claim first and then use it to prove what is asked: Towards, that end, consider

, cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! 🙂 )

In the given inequality:

we need to prove

consider where

LHS = , this is an AP with first term 2 and nth term and common difference 4; there are n factors “2”; hence, we so we get

; multiplying and dividing RHS of this by , we get the desired inequality. Remember the inequality is less than or equal to.

Problem 2:

Establish the Bernoulli inequality: If , then .

Solution 2:

Apply the binomial theorem, which in turn, is proved by mathematical induction ! 🙂

Problem 3:

For all natural numbers greater than or equal to 1, prove the following by mathematical induction:

Proof 3:

Let the given proposition be P(n).

Step 1: Check if P(1) is true. Towards that end:

and $latex RHS=2-\frac{1}{1}=2-1=1$ and hence, P(1) is true.

Step 2: Let P(n) be true for some , . That is, the following is true:

Add to both sides of above inequality, we get the following:

Now, the RHS in above is . We want this to be less than or equal to . Now, , , so what we have to prove is the following:

, that is, we want to prove that

, that is, we want , that is, we want , which is obviously true. QED.

Cheers,

Nalin Pithwa.