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Let $a_i,b_i,r_i,s_i$ be positive integers for $i\in\{1,2\}$. $r_i$ and $s_i$ are non-zero for $i\in\{1,2\}$.

Let $a=\left(\frac{1}{a_1},\frac{1}{a_2}\right), b=\left(\frac{1}{b_1},\frac{1}{b_2}\right),r=(r_1,r_2),s=(s_1,s_2)$, and $c=\left(\frac{1}{a_1+b_1},\frac{1}{a_2+b_2}\right).$

Let $\langle-,-\rangle $ denote the standard inner product on $\mathbb{R}^2$.

If $\langle a,r\rangle<1$ and $\langle b,s\rangle<1$, does that imply $\langle c,r+s\rangle\leq 1$?

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  • $\begingroup$ I think you can just do a bunch of tedious arithmetic to sort this out. $\endgroup$ Commented May 14, 2015 at 16:51
  • $\begingroup$ @ToddWilcox This is an interesting question, and is not that kind of easy I'm afraid. $\endgroup$
    – Daniel
    Commented May 14, 2015 at 16:54

1 Answer 1

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The question is equivalent to: $$ \text{If $\frac{r_{1}}{a_{1}} + \frac{r_{2}}{a_{2}} < 1$ and $\frac{s_{1}}{b_{1}} + \frac{s_{2}}{b_{2}} < 1$, is it true that $\frac{r_{1} + s_{1}}{a_{1} + b_{1}} + \frac{r_{2} + s_{2}}{a_{2} + b_{2}} < 1$?} \tag{1} $$ The answer is "no": With the "obvious" choices of numerators and denominators, we have $$ \frac{1}{2} + \frac{6}{13} < 1\text{ and } \frac{6}{8} + \frac{1}{8} < 1,\text{ but } \frac{7}{10} + \frac{7}{21} > 1. $$ (If it matters, this example was found by looking at a natural geometric interpretation of (1), guessing the result was false, looking at a "one-parameter family" of prospective counterexamples, and finding the smallest integer violating the conclusion.)

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  • $\begingroup$ I don't see why its a counterexample. This is an example where the equality occurs. So why this is a counterexample? My statement includes the equality. $\endgroup$ Commented May 15, 2015 at 8:37
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    $\begingroup$ My apology; I mis-read the conclusion as strict inequality. In that case,$$\frac{1}{2} + \frac{6}{13} < 1,\quad \frac{6}{8} + \frac{1}{8} < 1, \qquad \frac{7}{10} + \frac{7}{21} > 1$$is the "smallest" counterexample in the family I considered. $\endgroup$ Commented May 15, 2015 at 10:26

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