How to find maximum and minimum of $\frac{x+y+z}{ax+by+cz}$ where $0\leq x \leq y \leq z \leq 1$ for given positive real numbers $a,b,c$

Asked Mar 07 '14 at 03:41

Active Jan 23 '15 at 19:01

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How do I find the maximum and minimum of
$$\frac{x+y+z}{ax+by+cz}$$ where $0\leq x \leq y \leq z\leq 1$ for given positive real numbers $a,b,c$? I guess those are one of $\frac{3}{a+b+c}$ or $\frac{2}{b+c}$ or $\frac{1}{c}$, but can't prove it.

edited Jan 23 '15 at 19:01

Daniel W. Farlow

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asked Mar 07 '14 at 03:41

hanock

I haven't checked myself, but I am willing to bet that that function is monotonically increasing/decreasing in each of the three variables. If that's the case, extrema can only occur at endpoints, in which case you'll be done. – Ayesha Mar 07 '14 at 03:45
Something that might let you make some progress: if $v = (x, y, z)$, you've got $\frac{v\cdot(1, 1, 1)}{v\cdot (a, b, c)}$. The length of $v$ falls out, and then... you've got some work to do. – user24142 Jan 23 '15 at 19:14

How to find maximum and minimum of $\frac{x+y+z}{ax+by+cz}$ where $0\leq x \leq y \leq z \leq 1$ for given positive real numbers $a,b,c$

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