"Use mathematical induction to prove that if x > -1, then ( (1+x)^2 ) >= 1 + nx".

Question

Walk me through this, please

Answer 1

You have miswritten the problem. This is the Bernoulli inequality:

For x > -1, (1+x)^n >= 1 + nx

For induction, we prove for n = 1, assume for n = k, and prove for k + 1

1. n = 1 Show (1+x)^1 >= 1 + 1*x for x > -1: 1 + x = 1 + x, so 1 + x >= 1 + x for all x, including x > -1
2. Assume for n = k (1+x)^k >= 1 + kx for x > -1
3. For n = k + 1, (1+x)^n = (1+x)^(k+1) = (1+x)^(k)*(1+x) = (1+x)^(k) + (1+x)^k x

For x >= 0, (1+x) > 1, so (1+x)^k x > 1x = x

For -1 < x < 0, 0 < 1 + x < 1, so 0 < (1+x)^k < 1, and x < 0, so (1+x)^k x > x

Thus, for all x > -1, (1+x)^k x > x

Thus, From the induction step, (1+x)^(k+1) = (1+x)^(k) + (1+x)^k x >= 1+kx + (1+x)^k x >= 1+kx +x = 1 + (k+1)x

(1+x)^(k+1) >= 1 + (k+1)x

We have proven for n = 1, assumed for n = k, and proved for n = k+1, so our induction proof is complete.