Induction over n
true for n = 1
Assume true for n = k
that is, (1+x)^k >= kx+1
Then for n = k + 1
LHS = (1+x)^(k+1)
= (1+x) * (1+x)^k
>= (1+x) * (kx+1) (by induction)
= kx^2 + (k+1)x + 1
> 0 + (k+1)x + 1 (k is positive)
= (k+1)x + 1
So it is true for n = k + 1
Hence by induction the statement is true for positive n, x