i have a set of equations with inequalities and bounds to consider. I was able to reduce the original problem to the following form:
(1) $\sum_{i=1}^n x_i^2d_i <= 1$
(2) $\sqrt{\sum_{i=1}^n x_i^2} <= M$
So what we want to assume is that all $d_i$ are positive and that at most the $x_i^2$ "weighted" by the $d_i$s are summing up to 1. Clarification, the $d_i$s do not have to add up to one etc. despite they are acting like "weights":
$\sum_{i=1}^n d_i>0$
Key is to assume, that the $d_i$ values are kind of given and then regardless of the $x_i$s we want to things work like this:
Then, if the eq. (1) holds, then the eq. (2) should hold too. I.e. if the (1) equals one or less it should be possible to found a bound $M>0$ for the eq. (2), and actually that is what we want to show: if (1) is true, then (2) must be also for some $M<\infty$.
And as a sidenote, the case falls for the bound $M$ once not all $d_i$s are positive. This should happen, and is intuitively true, but what I want to show is that assuming some positive set of $d_i$s and given then that eq. (1) holds then the eq. (2) holds too for all $x_i$s satisfying eq. (1) given the $d_i$s.
Edit:
One possible solution, or at least a foothold. Since $d_i>0$, a square can be taken, let's define: $s_i^2=d_i$. Then let's put this into eq. (1): $\sum_{i=1}^n x_i^2d_i = \sum_{i=1}^n x_i^2s_i^2$
Looks familiar to Cauchy-Schawrz inequality, but that does not seem to be enough at least for now.
Found a solution, please check below and comment!
Got an answer. I considered deleting my question but I will keep it here so someone could verify my approach. Let's forget all the Cauchy-Schawrzs etc.
Let's build case from:
$\sum_{i=1}^n x_i^2d_i\le1$
Let's say $D=\min\{d_1,...,d_n\}$. Then, let's notice that the eq. (2) I gave, could be written as:
$\sqrt{\sum_{i=1}^n x_i^2} \le M \Rightarrow \sum_{i=1}^n x_i^2\le M^2$
Now the eq. (2) feels more managable. By combining some information, we can get:
$D\sum_{i=1}^n x_i^2 \le \sum_{i=1}^n x_i^2d_i \le 1$
From this, we can derive that:
$D\sum_{i=1}^n x_i^2\le 1$
Basic arithmetics:
$D\sum_{i=1}^n x_i^2\le 1 \Rightarrow \sum_{i=1}^n x_i^2\le 1/D$
Take the square root to arrive at the original definition of eq. (2) for LHS:
$\sum_{i=1}^n x_i^2 \le 1/D \Rightarrow \sqrt{\sum_{i=1}^n x_i^2} \le \sqrt{1/D}=1/ \sqrt{D}= 1/ \sqrt{\min\{d_i\}} =M$
Now we got some clearly defined bound that is $M>0$ strtictly, if all $d_i$ are positive, since if the minimum of $d_i$ is negative so will be also $1 / \sqrt{\min\{d_i\}}<0$ and that won't fit for the conditions of $M$.