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Let $x,x'\in \mathbb{R}^d$ with usual norm. \begin{equation} \frac{|x-x'|}{(1+|x|)(1+|x'|)} \leq \left|\frac{x}{|x|^2}-\frac{x'}{|x'|^2}\right| \end{equation} I have read this inequality, however, fail to prove. I appreciate if one can demonstrate how it follows.

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We need to prove that $$\left(\frac{|x-x'|}{(1+|x|)(1+|x'|)}\right)^2 \leq \left|\frac{1}{|x|^2}x-\frac{1}{|x'|^2}x'\right|^2$$ or $$\frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}\leq\frac{1}{|x|^2}+\frac{1}{|x'|^2}-\frac{2}{|x|^2|x'|^2}xx'$$ or $$(1+|x|)^2(1+|x'|)^2\geq|x|^2|x'|^2,$$ which is obvious.

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\begin{align*} \left|\frac{x}{|x|^2}-\frac{y}{|y|^2}\right|^2&=\frac{1}{|x|^2}+\frac{1}{|y|^2}-2\frac{x\cdot y}{|x|^2|y|^2}\\ &=\frac{|x|^2+|y|^2-2x\cdot y}{|x|^2|y|^2}\\ &=\frac{|x-y|^2}{|x|^2|y|^2}\\ &\ge\frac{|x-y|^2}{(1+|x|)^2(1+|y|)^2} \end{align*}

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  • $\begingroup$ Chrystomath.Second last line, correct? $\endgroup$ – Peter Szilas Jan 28 at 9:05

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