Uniformly distributed variables: what does the sum reveal. Say $U_1, U_2 \sim U(1,0)$ are independent uniformly randomly distributed variables on $[0, 1]$.
What lower bound $C$ should I enforce on their sum in order to believe (with a probability $p$) that at least one of the variables themselves is equal to at least $p$? In other words, what $C=C(p)$ solves the following
$$ 
P(U_1 \ge p \; \vee \; U_2 \ge p \mid U_1 + U_2 \ge C) = p.
$$

Geometrically this brings me to two distinct cases, namely:


*

*Case $C<p$ and the relation
$$
p = \frac{1-p^2}{1-C^2/2}.
$$

*Case $C\ge p$ and the relation
$$
p = \frac{1-p^2 - (C-p)^2}{1-C^2/2}.
$$



But what does it imply? How to interpret this intuitively? And where are the trivialities which undoubtedly arise when $C \notin [0,2]$?

Denote $A = \max \{ U_1, U_2\}$ and $B=U_1 + U_2$. And I suppose you want to find the smallest such $C$ for any $p$, thus the least information about $U_1 + U_2$.
First, suppose $C \le 0$, then we have $$\mathbb{P}(A \ge p \mid B \ge C) = \mathbb{P}(A \ge p) = 1- p^2.$$
If we want $\mathbb{P}(A \ge p)=p$, we must solve $1-p^2-p=0$, which yields $p = \frac{1}{2} ( \sqrt{5}-1)$. If we want $\mathbb{P}(A\ge p) \ge p$, then we have $p \le \frac{1}{2} ( \sqrt{5}-1)$.
So, if you have no information about the sum of $U_1 + U_2$, you can up to $p = \frac{1}{2}(\sqrt{5}-1)$ say that the maximum of $U_1$ and $U_2$ is at least $p$ with probability greater than or equal to $p$.
Now, if we take $0 \le C < p$, we have $$\mathbb{P}(A\ge p \mid B \ge C) \ge \mathbb{P}( A \ge p ).$$
So this is only interesting when $p \ge \frac{1}{2}( \sqrt{5} -1)$. This gives $$\mathbb{P}( A \ge p \mid B \ge C ) = \frac{ 1 -p^2}{1 - \frac{1}{2}C^2} \ge p$$
Solving yields 
\begin{eqnarray}
1 - p^2 &\ge& p - \frac{p}{2}C^2 \\
\frac{p}{2}C^2 &\ge& p + p^2 -1 \\
C^2 &\ge& -\frac{2}{p} + 2 + 2p
\end{eqnarray}
Since $C \le p$, we find that this can only work for approximately $p\le 0.68889$, which is bigger than $\frac{1}{2}(\sqrt{5}-1)$.
So, knowing that the sum greater than a constant, but not knowing if the sum is bigger than $p$, you can say that the maximum of $U_1$ and $U_2$ is at least $p$ with probability at least $p$.
So now the case $C \ge p$ remains, which has to be split into $1 \ge C \ge p$ and $2p \ge C \ge 1$, as $\mathbb{P}(B \ge C) = 2 - 2C + \frac{1}{2} C^2$ for $C \ge 1$, and when $C \ge 2p$, we have $\mathbb{P}(A \ge p \mid B \ge C) = 1$.
For $1 \ge C \ge p$, we find $C^2(1- \frac{1}{2}p) - 2pC -1+p +2p^2 \le 0$, plotting this numerical, we find that this holds for all $p \le 0.75$ approximately. ($x=C$ and $y=p$).

So the final case is $1 \ge C\ge 2p$, which a numerical approach yields that this doesn't hold.

