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

Question

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?

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• 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 $$ \max(a,c)+\max(b,c) = a + \max(b,c) = \max(a+b, a+c) \le \max(a+b, c). $$

Answer 1

Your method sounds correct.

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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 $c$.

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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.

Answer 2

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 $c>a+b$

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