Miklos Schweitzer 2014 - sum of reciprocal of lengths of intervals We let there be $k$ intervals within $[0,1]$. Prove that the sum of the reciprocals of the lengths of the intervals plus twice the sum of the reciprocals of the lengths of the nonempty intersection of any 2 intervals is at least $k^2$. 
 A: For any interval $I \subseteq [0,1]$, let, $\chi_I(x)$ be the indicator function for $I$ and $|I| = \int_0^1 \chi_I(x) dx$ be the corresponding length. 
Given any $k$ non-empty intervals $I_1, I_2, \ldots, I_k \subseteq [0,1]$,
consider the function defined by:
$$u(x) = \sum_{j=1}^k \frac{1}{|I_j|} \chi_{I_j}(x)$$
It is easy to see its average over $[0,1]$ is $k$:
$$\bar{u} = \int_0^1 u(x) dx = \sum_{j=1}^k \frac{1}{|I_j|} \int_0^1 \chi_{I_j}(x) dx = k$$
We have
$$
\int_0^1 \left( u(x) - \bar{u} \right)^2 dx \ge 0
\quad \implies \quad
\int_0^1 u(x)^2 dx \ge \bar{u}^2 = k^2
$$
Expanding the LHS, we find
$$\begin{align}
\text{LHS}
= & \sum_{j=1}^k \frac{1}{|I_j|^2} \int_0^1 \chi_{I_j}(x)^2 dx
 + 2\sum_{1 \le i < j \le k} \frac{1}{|I_i||I_j|}\int_0^1 \chi_{I_i}(x)\chi_{I_j}(x) dx\\
= & \sum_{j=1}^k \frac{1}{|I_j|^2} \int_0^1 \chi_{I_j}(x) dx
 + 2\sum_{1 \le i < j \le k} \frac{1}{|I_i||I_j|}\int_0^1 \chi_{I_i \cap I_j}(x) dx\\
= & \sum_{j=1}^k \frac{1}{|I_j|}
 + 2\sum_{\substack{1 \le i < j \le k,\\I_i \cap I_j \ne \emptyset}} \frac{|I_i \cap I_j|}{|I_i||I_j|}\\
\le & \sum_{j=1}^k \frac{1}{|I_j|}
+ 2\sum_{\substack{1 \le i < j \le k,\\I_i \cap I_j \ne \emptyset} } \frac{1}{|I_i \cap I_j|}\\
\end{align}$$
and what we want to show follows immediately.
