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 $>\frac{1}{3}x^2$.

Let, $f(x)=(1+x)^n-\frac{1}{2} n(n-1) x^2.$
So, $f'(x)=n(1+x)^{n-1}-n(n-1)x>0 \forall n>1$
Since,for $n>1$ and $x\ge 0,$ by A.M.-G.M. inequality we have,
$$\frac{1.(1+x)^{n-1}+(n-2).1}{n-1}>(1+x)$$
$$\implies (1+x)^{n-1}>(n-1)x$$
Therefore, $f$ is strictly increasing.
Hence we have,
$$f(x)>f(0)\implies (1+x)^n>\frac{1}{2}n(n-1) x^2$$
