Is there a tight upper bound on $\sum_{i=1}^n \sum_{j=1}^m \min(a\cdot i,b \cdot j)$ Is there a tight upper bound on $\sum_{i=1}^n \sum_{j=1}^m \min(a \cdot i,b \cdot j)$ for any $a,b \in \mathbb{R}^+$
For example one upper bound would be
\begin{align}
\sum_{i=1}^n \sum_{j=1}^m \min(a \cdot i,b \cdot j) \le \min \left(a \cdot \sum_{i=1}^n \sum_{j=1}^m i,b \cdot \sum_{i=1}^n \sum_{j=1}^m j \right)
\end{align}
Is there a better approach??
Thank you
Edit:
Note that w.l.o.g. we can set $a=1$. 
 A: This upper bound is surely not the tightest, but it's a cool approach i believe:
$$
\begin{align}
\sum_{i=1}^n \sum_{j=1}^m \min(a \cdot i,b \cdot j) \le & \sum_{i=1}^n \sum_{j=1}^m \frac{a \cdot i+b \cdot j}{2}=\\
&=\sum_{j=1}^m \sum_{i=1}^n \frac{a \cdot i}{2}+\sum_{i=1}^n \sum_{j=1}^m \frac{b \cdot j}{2}=\\
&=\frac{am}{2}\sum_{i=1}^n i+\frac{bn}{2}\sum_{j=1}^m j=\\
&=\frac{am}{2}\cdot\frac{n(n+1)}{2}+\frac{bn}{2}\cdot\frac{m(m+1)}{2}=\\
&=\frac{mn}{4}\Big[a(n+1)+b(m+1)\Big]
\end{align}
$$
so more concisely:
$$
\sum_{i=1}^n \sum_{j=1}^m \min(a \cdot i,b \cdot j) \le \frac{mn}{4}\Big[a(n+1)+b(m+1)\Big]
$$
by the way your upper bound can be written more concisely:
$$
\sum_{i=1}^n \sum_{j=1}^m \min(a \cdot i,b \cdot j) \le \frac{mn}{2}\min\Big(a(n+1),b(m+1)\Big)
$$
so you found a better upper bound apparently :D

mini-proof for: $\ \ \ \ \ \min(a,b)\le\frac{a+b}{2}$
given $a,b \in \mathbb{R}| a \le b$:
$$
min(a,b)=a\\
a\le b \iff a+a\le b+a \iff \frac{a+a}{2}\le \frac{a+b}{2} \iff a\le \frac{a+b}{2} \implies\\
\implies\min(a,b)=a\le \frac{a+b}{2}
$$
if $a>b$ just swap the letters in the proof :)

writing min(a,b) as an analytic function:
$$\min(a,b)= \frac{a+b-|a-b|}{2} = \frac{a+b-\sqrt{(a-b)^2}}{2} $$
