# Let $a,b,c \le 0$. Then $\max(a,c)+\max(b,c) \le \max(a+b, c)$

I'm reading a lecture note in which the author uses

Let $$a,b,c \le 0$$. Then $$\max(a,c)+\max(b,c) \le \max(a+b, c).$$

Below is my attempt. Is there other way to look at the problem and have a more direct solution?

• Assume $$a\le c$$ and $$b \le c$$. Notice that $$2c \le c$$, so $$\max(a,c)+\max(b,c) = 2c \le \max(a+b, c).$$
• Now consider the case at least $$a>c$$ or $$b>c$$. WLOG, we assume $$a>c$$. Notice that $$a+c \le c$$, so $$\max(a,c)+\max(b,c) = a + \max(b,c) = \max(a+b, a+c) \le \max(a+b, c).$$
• Really great @GarethMa! It's more direct. Could you post your comment as an answer? Commented May 19, 2022 at 15:57
• Ah sorry, I deleted my comment to post an answer because I just saw your question about "is there a more direct solution" Commented May 19, 2022 at 16:00

Your method sounds correct.

Another method I can think of is $$\max(a, c) + \max(b, c) = \max(a + b, a + c, b + c, 2c)$$, and noticing that each term is either at most $$a + b$$ or at most $$c$$.

Yet another method I can think of is writing $$a, b, c = -a', -b', -c'$$ where $$a', b', c' \geq 0$$. Then, your equation becomes

$$\max(-a', -c') + \max(-b', -c') \leq \max(-(a' + b'), -c') \\ \min(a', c') + \min(b', c') \geq \min(a' + b', c')$$

Which you can then apply the same $$LHS \geq \min(a' + b', a' + c', b' + c', 2c')$$ thing.

Since $$a,b,c≤0$$

$$a+b≤a$$ and $$a+b≤b$$

$$c≤a+b \iff c≤a$$ and $$c≤b \qquad \ldots \;(1)$$

So, $$\;c>a+b \iff c>a$$ or $$c>b \qquad \ldots \;(2)$$

• When $$c≤a+b$$

$$\max(a,c)+\max(b,c) =a+b =\max(a+b, c)\quad (\text{using } (1))$$

• When $$c>a+b$$

$$\max(a,c)+\max(b,c) =(a+c \text{ or } b+c \text{ or } c+c) \lt c=\max(a+b, c)\quad (\text{using } (2))$$