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Is it true that $$\sup_{x\in\mathbb{R}}\frac{\left|\left(1+s\right)+tx\right|+\left|\left(1+t\right)x+s\right|}{1+\left|x\right|+\left|s+tx\right|}\geq1$$ for all $s,t\in\mathbb{R}$? Is it also true that $$\sup_{x\in\mathbb{R}}\frac{\left|\left(1+s\right)x+t\right|+\left|\left(1+t\right)+sx\right|}{1+\left|x\right|+\left|sx+t\right|}\geq1 \ ?$$ Any answer to either of the two questions would be highly appreciated.

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There are troubles with the maximum: it may not exist at all! Take $s=-1$, $t=0$ in the 1st relation; the expression under max is $|x-1|/(2+|x|)$ which is strictly less than 1 for any $x\in\mathbb R$, but the supremum is 1. Please clarify your question. –  Wadim Zudilin Jun 15 '10 at 4:50
    
Wadim, thank you for pointing out. They should be "sup" instead of "max". –  user4606 Jun 15 '10 at 5:21
    
I've removed the "functional analysis" tag and added "inequalities". –  Ian Morris Jun 15 '10 at 9:17
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1 Answer 1

If I denote $$ f(s,t;x)=\frac{\left|\left(1+s\right)+tx\right|+\left|\left(1+t\right)x+s\right|}{1+\left|x\right|+\left|s+tx\right|}, $$ then for $t\ne0$ $$ f\left(s,t;-\frac{s}{t}\right)=\frac{1+|s/t|}{1+|s/t|}=1, $$ hence the supremum over all $x\in\mathbb R$ is at least 1. If $t=0$, then $$ f(s,0;x)=\frac{|1+s|+|x+s|}{1+|s|+|x|} $$ and $$ \lim_{x\to\pm\infty}f(s,0;x)=1, $$ so the supremum is (at least) 1 as well.

A similar substitution $x=-t/s$ works for the second supremum.

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