Show that $(1+x)^n\geq 1+nx+\Bigl(\frac{n(n-1)}{2}\Bigr)x^2$

Question

Show that $$(1+x)^n\geq 1+nx+\Bigl(\frac{n(n-1)}{2}\Bigr)x^2,\qquad x>0$$

I tried by induction

For $n=1$ $$(1+x)\geq1+nx$$ Assuming that $P(k)$ is true for $k\in \mathbb{N}$, so we need to show that for $k+1$ it is $$(1+x)^{k+1}\geq1+(k+1)x+\Bigl(\frac{(k+1)k}{2}\Bigr)x^2$$

So I think if $P(k)$ is true, I just need to multiply both sides by $(1+x)$ and show that is true for $(k+1)$, but if I do

$$(1+x)^n(1+x)\geq \Bigl[1+nx+\Bigl(\frac{n(n-1)}{2}\Big)x^2\Bigr](1+x)$$

a cubic term shows up, what makes no sense. Can anyone help me?

Answer 1

Following your calculation, we have \begin{align} &[1+nx+\Big(\frac{n(n-1)}{2}\Big)x²](1+x) \\ =\ &1 + \color{red}{nx} + \color{blue}{\left(\frac{n(n-1)}{2}\right)x^2} + \color{red}{x} + \color{blue}{nx^2} + \left(\frac{n(n-1)}{2}\right)x^3 \\ >\ &1 + (n + 1)x + \left\{\left(\frac{n(n-1)}{2}\right) + n\right\}x^2\\ =\ &1 + (n + 1)x + \left(\frac{(n+1)n}{2}\right)x^2 \end{align} where the third line is because the cubic term is greater than $0$.

Answer 2

Hint: Try to use the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem)

Answer 3

This can be proved by induction: $(1 + x)^{n+1} = (1+x)^n * (1 + x) \ge (1 + x*n + \frac{n*(n-1)*x^2}{2})*(1 + x) \ge 1 + x * (n + 1) + x^2 * n = 1 + x * (n + 1) + \frac{n*(n+1)*x^2}{2}$

Answer 4

If you can use derivatives, let $$ f_n(x)=(1+x)^n-1-nx-\frac{n(n-1)}{2}x^2 $$ which, by induction hypothesis is nonnegative for $x\ge0$.

Then, for $x>0$, \begin{align} f_{n+1}'(x) &=(n+1)(1+x)^n-(n+1)-n(n+1)x \\[6px] &=(n+1)\bigl((1+x)^n-1-nx\bigr)\\[6px] &\ge\frac{n(n^2-1)}{2}x^2\\[6px] &>0 \end{align} Thus $f_{n+1}$ is increasing. Since $f_{n+1}(0)=0$, we have the thesis.