Prove by induction this inequality

Question

Prove by induction that for $a>0, (1+a)^n>\frac{n(n-1)a^2}{2}$. I can prove it by Newton's Binomial but I need to prove it by induction.
Actually , I must add that I've already proved it but by using another induction (which not seems good enough) :
checking for n=1 is o.k. , assume for n that :
$a>0, (1+a)^n>\frac{n(n-1)a^2}{2}$
and need to prove for n+1 that : $$a>0, (1+a)^{n+1}>\frac{(n+1)n*a^2}{2}$$
proof :
$(1+a)^{n+1}=(1+a)^n(1+a)=(1+a)^n+(1+a)^n*a>\frac{n(n-1)a^2}{2}+(1+a)^n*a$
using a different induction (easy to prove) : $a>-1, (1+a)^n \geq 1+an$
we can write :
$\frac{n(n-1)a^2}{2}+(1+a)^n*a \geq \frac{n(n-1)a^2}{2} + (1+an)*a=\frac{n(n-1)a^2}{2} + a + n*a^2>$
$\frac{n(n-1)a^2}{2} + n*a^2 =\frac{(n(n-1)+2*n)a^2}{2}=$
$=\frac{(n^2+n)a^2}{2}=\frac{n(n+1)a^2}{2}$ and we've done .
Can anyone help please ? Thanks in advance.

Answer 1

we have \begin{eqnarray*} (1+a)^n&=&\sum^n_{k=0}C^k_na^k\\ &\geq& C_n^2a^2=\frac{n(n-1)a^2}{2}\\ \end{eqnarray*}