Given that $|a|=1$
$|b-2|=3$
$|c-5|=6 $ find the max value of$ |2a-3b-4c|$ where a,b,c are complex numbers.
I solved it like this
$|b-2|≤|b|+|-2|$
$3\le |b| + 2$
$1\le |b|$
$3\le |3b|$
Similarly i got
$4≤|4c|$
Then i wrote
$|2a-3b-4c|≤|2a|+|3b|+|4c|$
$|2a-3b-4c|≤2+3+4$
$|2a-3b-4c|≤9$
But 9 isn't the right answer.
On the other hand if I write $-|2|$ instead of $|-2|$ in
$|b-2|≤|b|+|-2|$
And similarly in the case of c as well
I get the answer as 61 which is right.
On
The reason your answer didn’t work was because you maximised $a,b$ and $c$ separately (that too incorrectly).
Now such an approach might have worked, if there was $+$ sign everywhere. Since we don’t, you have to solve it differently.
$$|2a-3b-4c|\le |2a|+|3b|+|4c|$$
Now maximise each term separately, for example $$|3b|\le 3|b-2+2|$$
$$\le 3||b-2|+2|$$ $$\le 3|(3+2)|$$ $$\le 15$$
Do the same for all the three terms.
Note,
$$|2a-3b-4c|\le|2a|+|3b|+|4c|$$ $$=2|a|+3|b-2+2|+4|c-5+5|$$ $$\le |2a|+3(|b-2|+2)+4(|c-5|+5)$$ $$=2\cdot 1+3\cdot(3+2)+4\cdot(6+5)=61$$
where the maximum value $61$ occurs at $a=-1$, $b=5$ and $c=11$.