I came up with this inequality, and I wanted to know if it has been said before. Also I want to ask if it can be proven in a easier way, because my proof is quite exhausting.
theorem:
Let $A$ be a geometric series, with a positive first term, and an odd number of terms. The mean of the terms with odd indexes, is greater or equal to the mean of terms with even indexes. $$\begin{cases}A=\{a_1,a_2,...a_{2n+1}\}\\a_1=c>0\\\frac{a_{k+1}}{a_k}=q\in\mathbb{R}\rightarrow a_k=cq^{k-1}\end{cases}\ \ \ \rightarrow\ \ \ \frac{c+cq^2+cq^4+...+cq^{2n}}{n+1}\geq\frac{cq+cq^3+...+cq^{2n-1}}{n}$$
my attempt of proof:
First of all we'll divide both sides by $c>0$, so it will not change what we have to prove. We can also multiply both sides by $\ n(n+1)>0$, and we'll have to prove: $$n(1+q^2+q^4+...+q^{2n})\geq(n+1)(q+q^3+...+q^{2n-1})$$
Moving all to the left side: $$nq^{2n}-(n+1)q^{2n-1}+nq^{2n-2}-(n+1)q^{2n-3}+...-(n+1)q+n\geq0$$
What we have here, is a polynomial with a degree of $2n$.
It's easy to see that $1$ is a root of this polynomial, so lets factor out $(q-1)$: $$(q\!-\!1)(nq^{2n-1}\!-\!q^{2n-2}\!+\!(n-1)q^{2n-3}-2q^{2n-4}+(n-2)q^{2n-5}-...-(n-1)q^2+q-n)\geq0$$ What did just happened? Well, the first coefficient must be $n$, so we have already $-n$ for $q^{2n-1}$, but we need $-(n+1)$, so we add $-1q^{2n-2}$, so now we have already $1$ for the next power, but we need $n$, so we add $(n-1)$ for the next one. And so further. The coefficients of the odd powers are going down from $n$ to $1$, and the coefficients of the even powers are going down from $-1$ to $-n$, which means that ones again $1$ is a root of this new polynomial inside the brackets. So we factor out $(q-1)$ ones again: $$(q-1)^2(nq^{2n-2}+(n-1)q^{2n-3}+2(n-2)q^{2n-4}+2(n-3)q^{2n-5}+3(n-3)q^{2n-6}+...)$$ What we see here is this; the first coefficient is $n$, so we have $-n$ for the second term, but we need only $-1$, so we add $(n-1)$, now we get for the third term $-(n-1)$, but we need $+(n-1)$, so we add $2(n-1)$, and so on...
Now we'll see that all coefficients inside the brackets are positive! We can see that for any odd power of the kind $2(n-k)-1$, the coefficient will be $k(n-k)$, and for any even power of the kind $2(n-k)$, the coefficient is $k(n-[k-1])$, and since the powers are going down as long as $n\geq k$, all coefficients are positive.
So the new polynomial is always positive, and it's multiplied by a square, the proof is complete.
Can anyone come up with a nicer proof?
Sure, maybe we can prove this a little more cleanly!
Let's show that for $c > 0$ and $q > 1$, $$ \frac{c+cq^2+cq^4+...+cq^{2n}}{n+1}\geq\frac{cq+cq^3+...+cq^{2n-1}}{n}. $$
Divide by $c$ and then rewrite it as $$ \frac{\left( \frac{q^{2n+2} - 1}{q^2 - 1} \right)}{n+1} \ge \frac{\left( q\frac{q^{2n} - 1}{q^2 - 1} \right)}{n} $$ equivalently, $$ nq^{2n+2} - n \ge (n+1) q^{2n+1} - (n+1)q. $$ i.e. $$ n(q-1)q^{2n+1} + n(q-1) \ge q(q^{2n} - 1) $$ i.e. $$ \frac{q(q^{2n} - 1)}{q^{2n+1} + 1} \le n. $$ But this is easy because $$ \frac{q(q^{2n} - 1)}{q^{2n+1} + 1} = \frac{q^{2n+1} - q}{q^{2n+1} + 1} < \frac{q^{2n+1} +1}{q^{2n+1} + 1} = 1 \le n. $$