How to write $\min\{a,k\}+\min\{b,k\}$ as one min?

Question

I want to simplify a solution to a problem I have where I used to write $$\min\{a,k\} + \min\{b,k\} \leq \min\{c,k\} + \min\{d,k\}.$$ I am wondering: is there a more simplified way of writing this line since $k$ is a common argument in all minimums?

Answer 1

Your expression will always equal $$\min\{a+b, a+k, k+b, 2k\}.$$ However, whether that is simpler than your expression is subjective. I would say the original is clearer.

Answer 2

If we don’t know which is smaller than other, we may use analytic approximation:

$$\lim_{x\rightarrow -\infty} \frac{\log(e^{ax} + e^{bx})}{x} = \min\{a,b\}$$

Now, in our case;

\begin{align} &=\lim_{x\rightarrow -\infty} \frac{\log(e^{ax} + e^{kx})}{x} + \lim_{x\rightarrow -\infty} \frac{\log(e^{bx} + e^{kx})}{x} \\ &=\lim_{x\rightarrow -\infty} \frac{\log((e^{ax} + e^{kx})(e^{bx} + e^{kx}))}{x} \\ &=\lim_{x\rightarrow -\infty} \frac{\log(e^{(a+b)x} + e^{(a+k)x}+ e^{(b+k)x}+ e^{(2k)x})}{x} \\ &=\min\{a+b, a+k, b+k, 2k\} \end{align}

This was already a result shown by @5xum. I was just explaining how this happened.