Proving the existence of infinitely many numbers with a given property in a non-increasing sequence.. $\textbf{Question:}$
Let $a_1, a_2, a_3,\cdots$   be a nonincreasing sequence of positive real numbers such that
$a_n \ge a_{2n} + a_{2n+1}$ for all $n \ge 1$.
Show that there exist infinitely many positive integers $m$ such that
$2m \cdot a_m > (4m − 3) \cdot a_{2m−1}.$
I haven't made any good progress in this problem.I noticed that,if $a_n=a_{n-1}$ infinitely many times,then it holds true.Any kind of hint or full solution both are appreciated.
 A: I think this works do tell me if you found an error
Notice that $2^{n+1} - 2^{n} = 2^{n}$ we will use this. Now let on contrary the above condition is true for only finitely many positive integers, thus there exist $ K $ such that for all $m \geq K$ we have
$$2m\cdot a_m \leq (4m-3)\cdot a_{2m-1}$$ Now consider the index of the sequence $(a)_N$ from
$(2^{r} \cdot m + 1)_{r \in N}$ for sufficiently large $m \geq K$. Now consider
$$2 \cdot  (2^{r} \cdot m + 1) \cdot a_{2^r \cdot m + 1} \leq (4 \cdot (2^r \cdot m + 1) -3) \cdot a_{2^{r+1} \cdot m + 1}$$
which gives
$$2 \cdot (2^{r} \cdot m + 1) \cdot a_{2^r \cdot m + 1} \leq (2^{r+2} \cdot m + 1) \cdot a_{2^{r+1} \cdot m +1}$$
hence
$$ \frac{a_{2^{r+1} \cdot m + 1}}{a_{2^r \cdot m + 1}} \geq \frac{2 \cdot (2^r \cdot m + 1)}{(2^{r+2} \cdot m +1)}$$
now consider the inequalities for $ r \in \{0, 1, 2, 3, \cdots , n\}$
we can multiply them to get
$$\frac{a_{2^{n+1} \cdot m + 1}}{a_{m+1}} \geq 2^{n+1} \cdot (m+1) \cdot (2m+1) \cdot \left( \frac{1}{(2^{n+1} \cdot m + 1) \cdot (2^{n+2} \cdot m + 1) } \right)$$
hence
$$ a_{2^{n+1} \cdot m + 1} \geq a_{m+1} \cdot (m+1) \cdot (2m+1) \cdot \left( \frac{1}{(2^{n+1} \cdot m + 1)} -  \frac{1}{(2^{n+2} \cdot m + 1)} \right) \cdot \frac{1}{m}$$
Let $$ p_{n + 1} \coloneqq \left( \frac{1}{(2^{n+1} \cdot m + 1)} -  \frac{1}{(2^{n+2} \cdot m + 1)} \right) \cdot  \frac{(m+1) \cdot (2m+1)}{m} $$
hence we have
$$ a_{2^{n} \cdot m + 1} \geq a_{m+1} \cdot p_{n}$$
now also from the first condition we have
$$a_{n} \geq a_{2n} + a_{2n+1}$$ putting $n = 2^{r-1} \cdot m$ we get
$$a_{2^{r-1} \cdot m} \geq a_{2^r \cdot m} + a_{2^r \cdot m + 1}$$
hence
$$a_{2^{r-1} \cdot m} - a_{2^r \cdot m } \geq a_{2^r \cdot m + 1}$$
summing both side from $r = 1 $ to $ r = n $ we get
$$\sum_{r = 1}^{n} a_{2^r \cdot m + 1} \leq a_{m} - a_{2^n \cdot m}$$
$$\Rightarrow a_m \geq \sum_{r = 1}^{n} a_{2^r \cdot m + 1} + a_{2^ n \cdot m} $$
$$\Rightarrow a_m \geq \sum_{r = 1}^{n} a_{2^r \cdot m + 1} + a_{2^ {n+1} \cdot m} $$
because $(a_n)$ is non increasing hence we have
$$a_{m} \geq \sum_{r = 1}^{n+1} a_{2^r \cdot m + 1} $$
$$a_{m} \geq \sum_{r = 1}^{n+1} a_{2^r \cdot m + 1} \geq \sum_{r = 1}^{n+1} (a_{m + 1} \cdot p_{r}) $$
giving
$$a_m \geq \frac{a_{m+1} \cdot (m+1) \cdot (2m+1)}{m} \cdot \left(\frac{1}{2m+1} - \frac{1}{2^{n+2}\cdot m + 1} \right)$$\
now as $n \rightarrow \infty$ we get
$$a_{m} \cdot m \geq a_{m+1 \cdot (m+1)}$$
$$\Rightarrow a_{m} \cdot m \geq a_{m+1 \cdot (m+1)} \geq a_{m+2} \cdot (m+2) \cdots \geq a_{2m-1} \cdot (2m-1)$$
Notice this happens only for sufficiently large $m \geq K$
$$2 \cdot a_{m} \cdot m \geq a_{2m-1} \cdot (4m-2) > a_{2m-1} \cdot (4m-3) $$
contradiction.
