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Let $a,b,c,d$ be constants in the interval $[-1,1]$. Define $$f(x,y)=\max\{|y-a|,1-b\}+\max\{1-x,1-y\}+\max\{|x-c|,1-d\}$$ for $ -1\le x\le 1, -1\le y\le 1.$ Prove, or disprove, that the minimum value of $f$ is $$\max\{2-b-c,2-a-d, 2-b-d\}.$$ Numerical evidence seems to show that this is true.

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That is true. Let $A=2-b-c, B=2-a-d, C=2-b-d$. Note that $f(x,y)\ge \max\{A,B,C\}$ for all $(x,y)$ in the domain. To prove that there exist $(x,y)$ in the domain such that $f(x,y)= \max\{A,B,C\}$ consider six cases: $A\le B\le C, A\le C\le B,\cdots$. For example in the case $C\le B\le A$, we have $$a\le b, c-d\le a-b.$$ If $1+c-d\le a+b-1$, we choose $x,y$ such that $$1+c-d\le x\le y \mbox{ and } a+b-1\le y\le 1+a-b,$$ and if $a+b-1\le 1+c-d$, we choose $x,y$ such that $$1+c-d\le x\le y\le 1+a-b.$$ Direct checking shows that $f(x,y)= \max\{A,B,C\}=A.$ Other cases are similar; in some cases $x=y=1$ suffices.

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