Here is the question: Prove that $(1-x)^n \geq 1-nx,~\forall n\in\mathbb{N}~and~x\in(0,1)$.
My proof by induction:
Base Case: when $n=1$, $(1-x)^1\geq 1-1\cdot x$
Induction Hypothesis: $\forall i \in [1,k],~(1-x)^i\geq 1-ix$
Inductive Steps: $(1-x)^{k+1}=(1-x)^k(1-x)\geq (1-kx)(1-x)=1-(k+1)x+kx^2>1-(k+1)x$
However, by this reasoning, I have only shown that $(1-x)^{k+1}>1-(k+1)x$ and have never shown that the equality.
Then I tried an alternative to show that, when $n=1$, the equality holds and then prove that the inequality holds for every natural number $n\geq2$. But I am not sure that the alternative reasoning is valid.
I know similar question has been posted before but in this one, $0<x<1$, so I cannot just reason that $1-(k+1)x+kx^2\geq 1-(k+1)x$.
Many thanks,
D.
Equality can only hold when $k=0$, when the $kx^2$ contributes nothing. This can be seen directly from your proof. The last inequality should be $\geq$ to account for the $k=0$ case.