A weird Inequality of the sum of two series I have an inequality which I believe is true but it seems too mysterious to prove it. Suppose $a_0,a_1,\cdots,a_{k-1}$ are positive integers such that $$a_0>\sum_{i=1}^{k-1}a_k $$
I want to prove that $$4\sum_{i=0}^{k-1}(k-i)^2a_i-\frac{1}{a_0} \left(\sum_{i=0}^{k-1}(k-i)a_i\right)^2 \geq k^2a_0$$
The reason why I believe this is true is that $\frac{1}{a_0}$ is a really small number which makes sure the second term on the left is not too big. Induction on $k$, though works on the base case $k=1$, is not very feasible since the square of the sum in the second term would be hard to handle. 
Since we are interested in the lower bound, we may substitute $a_0 =\sum_{i=1}^{k-1} a_k$ but this may not get any prettier either. Another feasible solution is to use Cauchy-Schwarz on the second term, but it will produce $k^3$ since it would involve the sum of squares. 
Here is the approach I'm taking right now, I assume $a_1$ dominates the series of $\sum_{i=1}^{k-1}a_k$, so we have the left term is larger than $$4\sum_{i=0}^{k-1}(k-i)^2a_i-\frac{1}{a_o}(ka_o+(k-1)a_0)^2= 4\sum_{i=0}^{k-1}(k-i)^2a_i-(2k-1)^2a_0$$ so we just need to prove $$4\sum_{i=0}^{k-1}(k-i)^2a_i \geq (5k^2-4k+1)a_0$$
At this stage, I feel this approach may be too aggressive since the left term contains $4k^2a_0$ but $5k^2a_0$ dominates the right term. I always forgot tricks in proving inequalities. Thank you for any suggestions.
 A: Wow, thanks buddy that was truly an egregious claim, super brainfart so thank you for pointing that out. And now we change our mind on the answer to the Q and try again lol (using $a_0$ to $a_k$ for easier notation):
$$4(\sum_{i=0}^{k}(k-i)^2a_i)-\frac{1}{a_0}(\sum_{i=0}^k(k-i)a_i)^2\geq k^2a_0$$ becomes
$$4a_0(\sum_{i=0}^{k}(k-i)^2a_i)-(\sum_{i=0}^k(k-i)a_i)^2\geq k^2a_0^2$$
and we see that
$$4a_0(\sum_{i=0}^{k}(k-i)^2a_i)-(\sum_{i=0}^k(k-i)a_i)^2=$$
Split the second term into the squares and non-squares:
$$4a_0(\sum_{i=0}^{k}(k-i)^2a_i)-[(\sum_{i=0}^k(k-i)^2a_i^2)+2(\sum_{0\leq i<j\leq k}(k-i)(k-j)a_ia_j)]\geq$$
In our middle term as $a_0>a_i$,
$$4a_0(\sum_{i=0}^{k}(k-i)^2a_i)-[(\sum_{i=0}^k(k-i)^2a_0a_i)+2(\sum_{0\leq i<j\leq k}(k-i)(k-j)a_ia_j)]=$$
$$3a_0(\sum_{i=0}^{k}(k-i)^2a_i)-2(\sum_{i=0}^{k-1}\sum_{j=i+1}^k(k-i)(k-j)a_ia_j)\geq$$
Then as for any $0\leq i\leq k-1$, $(k-i)a_0> (k-i)\sum_{j=i+1}^ka_j\geq \sum_{j=i+1}^k (k-j)a_j$, we get $(k-i)a_0 > \sum_{j=i+1}^k(k-j)a_j$ and so the above inequality is greater than the below:
$$3a_0(\sum_{i=0}^{k}(k-i)^2a_i)-2(\sum_{i=0}^{k-1}(k-i)a_i(k-i)a_0)=$$
$$3a_0(\sum_{i=0}^{k}(k-i)^2a_i)-2a_0(\sum_{i=0}^{k-1}(k-i)^2a_i)=$$
$$a_0(\sum_{i=0}^{k}(k-i)^2a_i)\geq k^2a_0^2$$
as surely
$$\sum_{i=1}^k(k-i)^2a_i\geq 0$$
Hopefully this time I didn't butcher like the last. Sorry again!
