Let $w, v \in \mathbb{R}^n$ with the extra condition $w_j \in (0,1), \sum\limits_{i = 1}^n w_j =1 \thinspace (\dagger)$
I wish to find the tightest upper bound $C$ such that for all $w,v \in \mathbb{R}^n$ such that $w$ satisfies condition $\dagger$:
$$\sum\limits_{j = 1}^n v_j^2 w_j \leq C\sum\limits_{j = 1}^n v_j^2$$
For example: $w = \begin{bmatrix} 0.5 \\ 0.1 \\ 0.4 \end{bmatrix}$
$$ v_1^2 0.5 + v_2^2 0.1 + v_3^2 0.4 \leq C(v_1^2+v_2^2+v_3^2)$$
We could take the obvious answer $C =1$, so whatever $w$ I am thrown at $C=1$ always guaranteed to satisfy my inequality. But is there a even tighter bound?
You could even take $C = 0.5$ in your example. In particular, $C = \max_j w_j$ works. Without additional conditions, I believe this is the tightest upper bound, since it is possible that $w_j = \frac1n$ and $v_i = v_j = v$ for all $i, j$. In this case, $C$ must be at least $\frac1n = \max_j w_j$.
If it has to hold for every $v, w$, then $C = 1$ is the best you can do. Let $v = (1, 0, 0, \ldots, 0)$ and $w = (1-\epsilon, \frac{\epsilon}{n-1}, \frac{\epsilon}{n-1}, \ldots, \frac{\epsilon}{n-1})$ for any $\epsilon > 0$. Sending $\epsilon \rightarrow 0$ means we can require $C$ be arbitrarily close to 1.
In particular, suppose $C < 1$ works. Then take $v = (1, 0, 0, \ldots, 0)$ and $w = (\frac{1+C}{2}, \frac{1-C}{2(n-1)},\ldots, \frac{1-C}{2(n-1)})$. It is clear that $\sum_{j=1}^{n} v_j^2 w_j = \frac{1+C}{2} > C = C\sum_{j=1}^n v_j^2$. This is a contradiction, so $C \ge 1$. But its obvious that $C = 1$ is a valid option.