Prove by induction on $n$ that when $x \gt 0$, $ (1+x)^n \ge 1+nx+\frac{n(n-1)}{2}x^2 \text{ for all positive integers } n. $

Question

Here's the problem:

Prove by induction on $n$ that when $x \gt 0$

$$ (1+x)^n \ge 1+nx+\frac{n(n-1)}{2}x^2 \text{ for all positive integers } n. $$

So, clearly the base case is true. Here's how far I've gotten in the induction step:

$ (1+x)^{n+1} \ge 1+(n+1)x + \frac{(n+1)[(n+1)-1]}{2}x^2 $

$ (1+x)(1+x)^n \ge 1+nx+x+\frac{n^2+n}{2}x^2 $

$ (1+x)(1+x)^n \ge 1+[1+n+\frac{1}{2}(n^2+n)x]x $

...and that's pretty much it. I'm looking for an opportunity to use a triangle inequality, or maybe even Bernoulli's inequality, but nothing's popping out to me. Any hints would be appreciated. Thanks.

Answer 1

To complete the induction

We have: $$(1+x)(1+nx+\frac{n(n-1)}{2}x^2)=1+(n+1)x+nx^2+\frac{n(n-1)}{2}x^2+\frac{n(n-1)}{2}x^3$$

and using the inequality $x>0$ and $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}x^2$ you can finish your work, indeed you get: $$(1+x)(1+nx+\frac{n(n-1)}{2}x^2)\geq 1+(n+1)x+\frac{n(n+1)}{2}x^2$$

which gives the induction step!

Other method $$(1+x)^n=\sum_{i=0}^n \dbinom k n x^k\geq \sum_{i=0}^3 \dbinom k n x^k=1+nx+\frac{n(n-1)}{2}x^2$$

Answer 2

Hint: Using binomial theorem $$(1+x)^n=\sum_{k=0}^{n}\dbinom{n}{k}x^k1^{n-k}=1+nx+\dbinom{n}{2}x^2+\underbrace{\sum_{k=3}^{n}\dbinom{n}{k}1^kx^{n-k}}_{\gt0, \,\text{ for }x>0}$$