Will the returned result of the function

$$\max\{\tfrac{1}{a}+\tfrac{1}{f}, \tfrac{1}{b}+\tfrac{1}{e}, \tfrac{1}{c}+\tfrac{1}{d}\}$$

return the same set $\{a,f\}$, $\{b,e\}$ or $\{c,d\}$ as the function

$$\min\{a+f, b+e, c+d\}\quad?$$

Assume all numbers are positive and real-valued.

In other words, if, for example, $$a+f < b+e\quad\text{ and }\quad a+f < c+d,$$ will it be true that $$\tfrac{1}{a}+\tfrac{1}{f} > \tfrac{1}{b}+\tfrac{1}{e}\quad\text{ and }\quad \tfrac{1}{a}+\tfrac{1}{f} > \tfrac{1}{c}+\tfrac{1}{d}\quad ?$$

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    $\begingroup$ Do you mean $\text{argmin}$ i.e. the arguments $a,b,\ldots$ that minimize the corresponding sums? $\endgroup$ – Listing Nov 30 '11 at 9:13

No. For example, let $a=c=d=f=1$, $b=\frac{1}{2}$, and $e=2$. Then $$\max\{\tfrac{1}{1}+\tfrac{1}{1},\tfrac{1}{2}+\tfrac{1}{\frac{1}{2}},\tfrac{1}{1}+\tfrac{1}{1}\}=\max\{2,\tfrac{5}{2},2\}=\tfrac{5}{2}$$ returns $\{b,e\}$, but $$\min\{1+1,2+\tfrac{1}{2},1+1\}=\min\{2,\tfrac{5}{2},2\}=2$$ returns either $\{a,f\}$ or $\{c,d\}$ (take your pick).

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  • $\begingroup$ Wow, I really screwed up the question. Editing it right now to reflect what I really meant. Your have the right answer for the wrong question (stupid me). $\endgroup$ – Tabgok Nov 30 '11 at 9:13
  • $\begingroup$ @Tabgok: No problem, happens to everyone :) I've now edited my answer to reflect your new question, if I've understood it correctly. $\endgroup$ – Zev Chonoles Nov 30 '11 at 9:24
  • $\begingroup$ Thank you so much! Spent a long time trying to figure this out analytically - looked for a counter-example but everything I tried worked. $\endgroup$ – Tabgok Nov 30 '11 at 9:43

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