Let n be a positive integer, and let x>=-1. Prove, using induction, that (1+x)^n>=1+nx.

(1+x)^n>=1+nx

Assuming that x >=  -1, Prove this is true for n = 1

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

1 + x  >= 1 + x 

And this is true

Assume that it's true for n =   k, where n is a positive integer...... that is......assume that

(1 + x)^k >= 1 + kx      is true

Prove that it's true for n = k + 1

(1 + x)^(k + 1)  >=  1 + (k + 1)x

(1 + x)^(k + 1)  >= 1 + kx + x

(1 + x)^k * (1 + x) >=  ( 1 + kx) + x

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

( 1 + x)^k + x + x^2 >= (1 + kx) + x

(1 + x)^k + x^2 >= (1 + kx) +  0         (1)

And we assumed that (1 +x)^k >= (1 + kx)

And for all x, x^2 >= 0

So   (1)  is true

Then

(1+x)^n>=1+nx  is true for x >= -1 and n is a positive integer