Analysis: Proving the equality $(1+x)^n \geq \frac{1}{4}n^2x^2$.

Question

I'm trying to prove the inequality $(1+x)^n \geq \frac{1}{4}n^2x^2$, for $n \in \mathbb{N}, n \geq 2$ and $x \in \mathbb{R}, x \geq 0$.

What I have so far:

We prove by induction. Basis step: n = 2.

$$(1+x)^2 \geq \frac{2^2}{4}x^2$$ $$(1+x)^2 \geq x^2.$$

Which holds true for $n =2$. Inductive step: n = k + 1.

\begin{align} (1+x)^{k+1} &= (1+x)^k(1+x) \\ & \geq \frac{1}{4}n^2x^2(1+x) \\ & \geq \frac{1}{4}n^2(x^2 + x^3) \\ & \vdots \\ & = \frac{1}{4}(n+1)^2x^2 \end{align}

I can't seem to see how we can get $\frac{1}{4}(n+1)^2x^2$. I've tried to factor out a $nx^2$ or something along those lines, like proving the Bernoulli inequality, but to no avail.

Answer 1

$$ (1+x)^n= \sum_{k=0}^n \binom{n}{k} x^k \geq \binom{n}{2}x^2 \geq \frac{n(n-1)}{2}x^2 $$ When $n \geq 2$, $$ 2n \geq 4 \implies 4n-4 \geq 4n -2n \implies 4(n-1) \geq 2n \implies \frac{n-1}{2} \geq \frac{n}{4} $$ Which implies that $\frac{n(n-1)}{2} > \frac{n^2}{4}$. Hence,

$$ (1+x)^n \geq \frac{n(n-1)}{2}x^2 \geq \frac{1}{4}n^2x^2 $$