# Finding the minimum number $m$ so that the inequality $|5x+11y+17z|≤m(|x|+|y|+|z|)$ is true

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 $$x+y+z=0$$

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 $$m=6$$.

• Why we have m<=6 ? – asv Jun 29 '19 at 16:30
• Because the line above that shows that if you replace $m$ with $6$ in the condition, then the condition is satisfied. Therefore the minimum $m$ for which the condition is satisfied must be smaller or equal than 6. – Maurizio Moreschi Jun 29 '19 at 16:44
• Thank you very much – asv Jun 29 '19 at 17:46
• Why 6|x|+6|z|≤6(|x|+|y|+|z|) shows that m≤6? Why did you put 6 instead of m? – Futurballa Jun 30 '19 at 13:22
• @Futurballa What I prove in the first part of the answer is that, for any $x,y,z\in \mathbb{R}$ such that $x+y+z=0$, one has $|5x+11y+17z|\le 6(|x|+|y|+|z|)$. Therefore 6 is a number with the property you are requiring. Since you want $m$ to be the smallest number satisfying such property, you have $m\le 6$. – Maurizio Moreschi Jun 30 '19 at 14:08