If $f,g\in C[a,b]$, then is $\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$ true? 
If $f,g\in C[a,b]$, then is $$\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$$ true?

Intuitively, this feels right. Also, I think that $$\min_{a\le t\le b} |f(t) + g(t)| \ge \min_{a\le t\le b} |f(t)| + \min_{a\le t\le b} |g(t)|$$ must be true. What if we remove the absolute values, i.e. we talk about
$$\max_{a\le t\le b} f(t) + g(t) \le \max_{a\le t\le b} f(t) + \max_{a\le t\le b} g(t)$$ and
$$\min_{a\le t\le b} f(t) + g(t) \ge \min_{a\le t\le b} f(t) + \min_{a\le t\le b} g(t)$$
Do any of these hold? I'm not able to come up with formal methods to prove/disprove these, and I'd appreciate any help! I ended up using the very first inequality in a proof, so I hope it is right. About the rest, I am only speculating, but I would love to know how to approach inequalities like these. Thanks a lot!
P.S. $C[a,b]$ denotes the set of continuous functions on $[a,b]\subset \mathbb R$.
 A: Yes, all you wrote is correct except for the second inequality (with the $\min$'s and the absolute values). To see e.g. $$\max_{a\leq t\leq b}|f(t)+g(t)|\leq\max_{a\leq t\leq b}|f(t)|+\max_{a\leq t\leq b}|g(t)|$$
note that for all $s\in [a,b]$ we have $$|f(s)+g(s)|\leq |f(s)|+|g(s)|\leq\max_{a\leq t\leq b}|f(t)|+\max_{a\leq t\leq b}|g(t)|$$
Now taking the maximum over all $s\in [a,b]$ we get the inequality above. The other inequalites can be proven in the same way.
Notice that the inequality may be strict since $f$ and $g$ can attain their maxima at different points, consider e.g. $f(t)=t, g(t)=1-t$ on $[0,1]$.
To see why the second inequality fails, take e.g. $f=-g$.
A: It has already been pointed out in leoli1's answer that the second inequality is wrong. A correct lower bound for $\min |f(t) + g(t)| $ can be obtained by using the “reverse triangle inequality”:
$$
|f(t) + g(t)| \ge |f(t)| - |g(t)| = |f(t)| + (- |g(t)|)
$$
implies
$$
\min_{a\le t\le b} |f(t) + g(t)| \ge \min_{a\le t\le b} |f(t)| + \min_{a\le t\le b} (-|g(t)|) =  \min_{a\le t\le b} |f(t)| - \max_{a\le t\le b} |g(t)| \, .
$$
