<s>[INST] 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.
 [/INST] In fact, that identity is useful. In details, by AM-GM inequality, $$ \left(n-3\right)\left(1+a^n\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ and $$ \left(n-3\right)\left(a+a^{n-1}\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ Thus, $$ LHS = \frac{1+a+a^{n-1}+a^{n}}{a^2+\cdots+a^{n-2}}+1 \ge \frac{4}{n-3}+1 =RHS $$
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