<s>[INST] If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$ If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$. 
Is there a way to prove it without using the Binomial Theorem? Is it possible to use the Bernoulli's Inequality to prove it? If yes, please show me. 
I tried to prove it by induction on $n$, where $x>0$ is fixed. But it doesn't let me reach the conclusion. Here's what I did:

Fix $x>0$. We instead prove that
  \begin{equation*}
\frac{2(1+x)^n}{n(n-1)}> x^2
\end{equation*}
  for all $n>1$. Clearly, this is true for $n=2$. Assume this statement holds for $n=k$, for some $k\geq 2$. We have
  \begin{equation*}
\frac{2(1+x)^{k+1}}{(k+1)k}=\frac{2(1+x)^k}{k(k-1)}\frac{(x+1)(k-1)}{k+1}>\frac{(x+1)(k-1)}{k+1}x^2.
\end{equation*}
  Since $y\mapsto \frac{y-1}{y+1}$ is increasing on $\mathbb{R}\setminus \{-1\}$, we get $\frac{y-1}{y+1}\geq \frac{1}{3}$ for all $y\geq 2$. Therefore,
  \begin{equation*}
\frac{(x+1)(k-1)}{k+1}x^2\geq \frac{1}{3}(x+1)x^2>\frac{1}{3}x^2.
\end{equation*}

As you can see, I can not reach $>x^2$ instead of $>\frac{1}{3}x^2$.
 [/INST] As an alternative we can proceed by induction for the stronger
$$(1+x)^{n}>1+nx+\frac12n(n-1)x^2>\frac12n(n-1)x^2$$
that is for the induction step
$$(1+x)^{n+1}=(1+x)(1+x)^{n}\stackrel{Ind.Hyp.}>1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3>$$$$\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$
therefore we need to prove that
$$1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$
which is true indeed
$$\frac12n(n-1)x^2+nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12\left(n(n-1)+2n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12\left(n^2+n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12(n+1)nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12n(n-1)x^3>0$$
for $n>1$.
</s>