I'm trying to solve the following problem, but really have no idea where to start.
Find the minimum number $m ∈ R$ so that the inequality
$|5x+11y+17z|≤m(|x|+|y|+|z|)$
is true for every $x, y$ and $z$, with $x, y, z ∈ R$ and $x+y+z=0$
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityI'm trying to solve the following problem, but really have no idea where to start.
Find the minimum number $m ∈ R$ so that the inequality
$|5x+11y+17z|≤m(|x|+|y|+|z|)$
is true for every $x, y$ and $z$, with $x, y, z ∈ R$ and $x+y+z=0$
Since $x+y+z=0$ by assumption, we have $$5x+11y+17z=(5x+11y+17z)-11(x+y+z)=-6x+6z.$$ Therefore $$ |5x+11y+17z|=|-6x+6z|\le 6|x|+6|z|\le 6(|x|+|y|+|z|).$$ This shows that $m\le 6$.
On the other hand, if $y=0$, then $x=-z$, and the condition becomes $$ |12z|\le 2m|z| \quad \forall z\in \mathbb{R}.$$
Therefore $m\ge 6$, and thus $m=6$.