Can we simplify the convex sum of minima?

Question

Let $k\in\mathbb N$ and $\lambda_i,\mu_i,a_i,b_i\ge0$ for $i\in\{1,\ldots,k\}$ with $\sum_{i=1}^k\lambda_i=\sum_{i=1}^k\mu_i=1$. Can we simplify the expression $$ \sum_{i=1}^k\lambda_i\sum_{j=1}^k\mu_j\min(a_i,b_j)\tag1? $$ We may note that $2\min(a,b)=a+b-|a-b|$ and $2\max(a,b)=a+b+|a-b|$ for all $a,b\in\mathbb R$.

Answer 1

We have the trivial bounds $$ \min\{ \min(a_i, b_j) \;|\; 1 \le i,j \le k \} \le (1) \le \max\{ \min(a_i, b_j) \;|\; 1 \le i,j \le k \}.$$ These bounds are already sharp by choosing $\lambda_i = \mu_j = 1$ for $(i,j)$ attaining the min on the left (or the max on the right) and all other coefficients equal to $0$.