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


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 $m=6$.

  • $\begingroup$ Why we have m<=6 ? $\endgroup$ – asv Jun 29 at 16:30
  • $\begingroup$ 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. $\endgroup$ – Maurizio Moreschi Jun 29 at 16:44
  • $\begingroup$ Thank you very much $\endgroup$ – asv Jun 29 at 17:46
  • $\begingroup$ Why 6|x|+6|z|≤6(|x|+|y|+|z|) shows that m≤6? Why did you put 6 instead of m? $\endgroup$ – Futurballa Jun 30 at 13:22
  • $\begingroup$ @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 $m\le 6$. $\endgroup$ – Maurizio Moreschi Jun 30 at 14:08

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