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Can anyone show that $$\eqalign{&\max\left(\left|x_{1}-y_{1}+x_{2}-y_{2}\right|,\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}-2\left(x_{1}-x_{2}\right)\left(y_{1}-y_{2}\right)\cos2t}\right)\cr\qquad&\ge\left|\left|x_{1}\right|-\left|y_{1}\right|\right|+\left|\left|x_{2}\right|-\left|y_{2}\right|\right| \cr}$$ for all real numbers $x_{1}$, $x_{2}$, $y_{1}$, $y_{2}$, $t$? Thanks a lot.

[Edit:] In a comment and in a further, now deleted question, the OP asked this follow-up question:

Does the inequality hold if we assume $|x_1|\ge|x_2|$ and $|y_1|\ge|y_2|$? In fact, does

$$\max\left(\big|x_{1}-y_{1}+x_{2}-y_{2}\big|,\big|\left|x_{1}-x_{2}\right|-\left|y_{1}-y_{2}\right|\big|\right)\ge\big|\left|x_{1}\right|-\left|y_{1}\right|\big|+\big|\left|x_{2}\right|-\left|y_{2}\right|\big|$$

hold under those constraints?

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    Originally, the display spilled over the right margin, and everything to the right of the inequality sign was obscured by the "tagged" list. I edited to fix this. You've been having a rollback war with yourself over this. What's going on?2012-01-18
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    @Gerry: A civil rollback war, so to speak :-)2012-01-18

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