Using mathematical induction, prove the inequality
(1+a)^n >= 1+na for all a>-1 and all n
The Attempt at a Solution
1+a >= 1+a. the inequality holds.
I am struggling with the inductive step.
by working backward, i multiply both sides by (1+a) and this gives me
(1+a)^(k+1) >= (1+ka)(1+a)
all i can think of at this point is that 1+a >= 0, but i don't think this helps me.
working the other way, starting with n=k+1, i find
would is be enough to state that (1+a)> a and thus the inequality holds?
thanks in advance