Inequality related to a bijection $xmapsto |x|^{-2}x$












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$begingroup$


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.






inequality inner-product-space absolute-value







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    0












    $begingroup$


    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.






    inequality inner-product-space absolute-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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.






      inequality inner-product-space absolute-value







      share|cite|improve this question











      $endgroup$




      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.






      inequality inner-product-space absolute-value




      inequality inner-product-space absolute-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 10:38









      Michael Rozenberg

      109k1896201




      109k1896201










      asked Jan 28 at 4:58









      Jo'Jo'

      306110




      306110






















          2 Answers
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          $begingroup$

          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
          $$frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}leqfrac{1}{|x|^2}+frac{1}{|x'|^2}-frac{2}{|x|^2|x'|^2}xx'$$ or
          $$(1+|x|)^2(1+|x'|)^2geq|x|^2|x'|^2,$$ which is obvious.






          share|cite|improve this answer











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            0












            $begingroup$

            begin{align*}
            left|frac{x}{|x|^2}-frac{y}{|y|^2}right|^2&=frac{1}{|x|^2}+frac{1}{|y|^2}-2frac{xcdot y}{|x|^2|y|^2}\
            &=frac{|x|^2+|y|^2-2xcdot y}{|x|^2|y|^2}\
            &=frac{|x-y|^2}{|x|^2|y|^2}\
            &gefrac{|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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            2 Answers
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            2 Answers
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            $begingroup$

            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
            $$frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}leqfrac{1}{|x|^2}+frac{1}{|x'|^2}-frac{2}{|x|^2|x'|^2}xx'$$ or
            $$(1+|x|)^2(1+|x'|)^2geq|x|^2|x'|^2,$$ which is obvious.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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
              $$frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}leqfrac{1}{|x|^2}+frac{1}{|x'|^2}-frac{2}{|x|^2|x'|^2}xx'$$ or
              $$(1+|x|)^2(1+|x'|)^2geq|x|^2|x'|^2,$$ which is obvious.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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
                $$frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}leqfrac{1}{|x|^2}+frac{1}{|x'|^2}-frac{2}{|x|^2|x'|^2}xx'$$ or
                $$(1+|x|)^2(1+|x'|)^2geq|x|^2|x'|^2,$$ which is obvious.






                share|cite|improve this answer











                $endgroup$



                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
                $$frac{|x|^2+|x'|^2-2xx'}{(1+|x|)^2(1+|x'|)^2}leqfrac{1}{|x|^2}+frac{1}{|x'|^2}-frac{2}{|x|^2|x'|^2}xx'$$ or
                $$(1+|x|)^2(1+|x'|)^2geq|x|^2|x'|^2,$$ which is obvious.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 28 at 5:28

























                answered Jan 28 at 5:12









                Michael RozenbergMichael Rozenberg

                109k1896201




                109k1896201























                    0












                    $begingroup$

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






                    share|cite|improve this answer











                    $endgroup$













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
















                    0












                    $begingroup$

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






                    share|cite|improve this answer











                    $endgroup$













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














                    0












                    0








                    0





                    $begingroup$

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






                    share|cite|improve this answer











                    $endgroup$



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







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 29 at 7:22

























                    answered Jan 28 at 8:05









                    ChrystomathChrystomath

                    1,838513




                    1,838513












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


















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
















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




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


















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