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Is it true that $$\inf_{t,x,y\in\mathbb{R}}\max\left(\left|x-1\right|+\left|tx-1\right|,\left|y-1\right|+\left|ty-2\right|\right)=1/3?$$

Thanks.

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    $\begingroup$ Why do you ask and where did you get this inequality? Please motivat the question better. I'm also going to re-tag this question a bit, as some of the tags are quite obviously not applicable to this question. $\endgroup$ Nov 18, 2010 at 19:15
  • $\begingroup$ Note that taking $x=2/3$, $t=3/2$, $y=4/3$ shows the $\le$ direction. $\endgroup$ Nov 18, 2010 at 23:00
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    $\begingroup$ Unless there is something of particular interest about these expressions, I would call this "too localized" for MO. $\endgroup$ Nov 18, 2010 at 23:17
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    $\begingroup$ Please see the "how to ask page". $\endgroup$
    – S. Carnahan
    Nov 19, 2010 at 2:52

1 Answer 1

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Brute force should suffice here. Write $F(x,y,t) = \max\left(\left|x-1\right|+\left|tx-1\right|,\left|y-1\right|+\left|ty-2\right|\right)$. A moment's thought shows $F \ge 1/3$ outside the cube $[2/3, 4/3] \times [2/3,4/3] \times [1/2,2]$. You can divide this cube into simplices according to the signs of $x-1$, $tx-1$, $y-1$, $ty-2$; then in each simplex you are optimizing linear functions with respect to linear constraints, which is easy if tedious (assign it as a homework in a multivariable calculus course if necessary).

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  • $\begingroup$ +1 for the parenthetical. $\endgroup$ Nov 18, 2010 at 23:25

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