Geometric progression in an inequality Problem: Show that if $a>0$ and $n>3$ is an integer then $$\frac{1+a+a^2 \cdots +a^n}{a^2+a^3+ \cdots a^{n-2}} \geq \frac{n+1}{n-3}$$ 

I am unable to prove the above the inequality. 
I used the geometric progression summation formula to reduce it to proving $\frac{a^{n+1}-1}{a^2(a^{n-3}-1)} \geq \frac{n+1}{n-3}$.
Also writing it as $$\frac{1+a+a^2 \cdots +a^n}{n+1} \geq \frac{a^2+a^3+ \cdots a^{n-2}}{n-3}$$
 seems to suggest that some results on mean-inequalities can be used but I can't figure out what that is.

The inequality holds trivially for $a = 1$, and it is invariant
under the substitution $a \to 1/a$. Therefore it suffices to
prove the inequality for $\mathbf{a > 1}$.
The hyperbolic sine is a convex function on $[0, \infty)$,
so for $a > 1$ the function 
$$
   f(x) = 2 \sinh \bigl(\log a \cdot \frac x2 \bigr) 
= a^{x/2} - a^{-x/2}
= \frac{a^x - 1}{a^{x/2}} \quad (x \ge 0)
$$
is also convex, and therefore
$$
 \frac{f(x) - f(0)}{x-0} = \frac{a^x - 1}{x \, a^{x/2}}
$$
is increasing. It follows that for $0 \le x \le y$
$$
   \frac{a^y - 1}{y \, a^{y/2}} \ge \frac{a^x - 1}{x \, a^{x/2}} 
$$
which is equivalent to
$$
  \frac{a^y-1}{a^x-1} \ge a^{(y-x)/2} \frac yx \, .
$$
The desired inequality follows as a special case for $x = n-3$
and $y = n+1$, but we have shown that the inequality can be
generalized to arbitrary positive real numbers as exponents.
