Proof by induction: $(1+\alpha)^n\ge 1+n\alpha +\frac{n(n-1)}{2}\alpha ^2$

Question

so I have this problem. It asks me to prove an expression by induction.

Let $n$ be a positive integer, and $\alpha$ any nonnegative real number. Prove by induction that$$(1+\alpha)^n\ge 1+n\alpha +\frac{n(n-1)}{2}\alpha ^2$$

So I let $G(n) = (1+\alpha)^n\ge 1+n\alpha +\frac{n(n-1)}{2}\alpha ^2$. I showed the initial case when $n=0$ and got $1=1$. I then assume $G(k)$ is true $G(k) = (1+\alpha)^k\ge 1+k\alpha +\frac{k(k-1)}{2}\alpha ^2$. I then try to prove $G(k+1)$ is also true. $G(n) = (1+\alpha)^{k+1}\ge 1+(k+1)\alpha +\frac{(k+1)k}{2}\alpha ^2$. However, this is where I am stuck. I'm not sure where to go from here. Thanks.

Answer 1

You have

$$\begin{align} (1+\alpha)^{k+1}&=(1+\alpha)(1+\alpha)^k\\\\ &\ge (1+\alpha)\left(1+k\alpha+\frac{k(k-1)}{2}\alpha^2\right)\\\\ &=1+(k+1)\alpha+\frac{k(k+1)}{2}\alpha^2+\frac{k(k-1)}{2}\alpha^3\\\\ &\ge 1+(k+1)\alpha+\frac{k(k+1)}{2}\alpha^2 \end{align}$$

since $\alpha\ge 0$ and $k\ge 1$.

Answer 2

Here's a proof by induction showing that the inequality is true for any fixed size initial part of the binomial theorem.

Theorem: For fixed $m$ and $n \ge m$, if $x \ge 0$ then $(1+x)^n \ge \sum_{k=0}^m x^k\binom{n}{k} $.

Initial case: Expand $(1+x)^m$ manually. You will get $(1+x)^m =\sum_{k=0}^m x^k\binom{n}{k} $.

Suppose true for $n$. Then

$\begin{array}\\ (1+x)^{n+1} &=(1+x)(1+x)^{n}\\ &\ge(1+x)\sum_{k=0}^m x^k\binom{n}{k}\\ &=\sum_{k=0}^m x^k\binom{n}{k}+\sum_{k=0}^m x^{k+1}\binom{n}{k}\\ &=\sum_{k=0}^m x^k\binom{n}{k}+\sum_{k=1}^{m+1} x^{k}\binom{n}{k-1}\\ &=1+\sum_{k=1}^m x^k(\binom{n}{k}+\binom{n}{k-1})+x^{m+1}\binom{n}{m}\\ &=1+\sum_{k=1}^m x^k\binom{n+1}{k}+x^{m+1}\binom{n}{m}\\ &=\sum_{k=0}^m x^k\binom{n+1}{k}+x^{m+1}\binom{n}{m}\\ &\ge\sum_{k=0}^m x^k\binom{n+1}{k}\\ \end{array} $