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.