$||x|-|y||leq|x-y|Rightarrow |x-y|geq|x|-|y|$ and $|x+y|geq|x|-|y|$.












0












$begingroup$


I think I can prove the inequality but in order to do so I Need to understand whether if



$|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$



My proof would be then



$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$



And then one can choose for $y$ its negative value and would get



$||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$



If my idea is Right please help me to prove $(*)$



Otherwise I would like a hint so I can find it out myself






inequality absolute-value







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I think I can prove the inequality but in order to do so I Need to understand whether if



    $|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$



    My proof would be then



    $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$



    And then one can choose for $y$ its negative value and would get



    $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$



    If my idea is Right please help me to prove $(*)$



    Otherwise I would like a hint so I can find it out myself






    inequality absolute-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I think I can prove the inequality but in order to do so I Need to understand whether if



      $|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$



      My proof would be then



      $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$



      And then one can choose for $y$ its negative value and would get



      $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$



      If my idea is Right please help me to prove $(*)$



      Otherwise I would like a hint so I can find it out myself






      inequality absolute-value







      share|cite|improve this question











      $endgroup$




      I think I can prove the inequality but in order to do so I Need to understand whether if



      $|a|>|b|$ then $|a|> b$ and $|a| > - b (*)$



      My proof would be then



      $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x-y|$



      And then one can choose for $y$ its negative value and would get



      $||x|-|y||leq|x-y|Rightarrow |x|-|y|leq|x+y|$



      If my idea is Right please help me to prove $(*)$



      Otherwise I would like a hint so I can find it out myself






      inequality absolute-value




      inequality absolute-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 10:25









      Michael Rozenberg

      111k1897201




      111k1897201










      asked Jan 8 at 8:36









      RM777RM777

      38312




      38312






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.



          Hence $|a| > |b| ge b$, that is we have $|a| > b$.



          Similarly for $-b$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Because by the triangle inequality



            $$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
            $$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
            $$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
            $$|x+y|geq|x|-|y|.$$






            share|cite|improve this answer











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              2 Answers
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              active

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              active

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              active

              oldest

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              2












              $begingroup$

              We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.



              Hence $|a| > |b| ge b$, that is we have $|a| > b$.



              Similarly for $-b$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.



                Hence $|a| > |b| ge b$, that is we have $|a| > b$.



                Similarly for $-b$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.



                  Hence $|a| > |b| ge b$, that is we have $|a| > b$.



                  Similarly for $-b$.






                  share|cite|improve this answer









                  $endgroup$



                  We know that $|b|=max{b, -b}$, that is $|b| ge b$ and $|b| ge -b$.



                  Hence $|a| > |b| ge b$, that is we have $|a| > b$.



                  Similarly for $-b$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 8:38









                  Siong Thye GohSiong Thye Goh

                  104k1468120




                  104k1468120























                      0












                      $begingroup$

                      Because by the triangle inequality



                      $$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
                      $$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
                      $$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
                      $$|x+y|geq|x|-|y|.$$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Because by the triangle inequality



                        $$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
                        $$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
                        $$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
                        $$|x+y|geq|x|-|y|.$$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Because by the triangle inequality



                          $$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
                          $$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
                          $$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
                          $$|x+y|geq|x|-|y|.$$






                          share|cite|improve this answer











                          $endgroup$



                          Because by the triangle inequality



                          $$|x-y|+|y|geq|x-y+y|geq|x|,$$ which gives
                          $$|x-y|geq|x|-|y|$$ and by the triangle inequality again:
                          $$|x+y|+|y|=|x+y|+|-y|geq|x+y-y|=|x|,$$ which gives
                          $$|x+y|geq|x|-|y|.$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 8 at 9:35

























                          answered Jan 8 at 8:40









                          Michael RozenbergMichael Rozenberg

                          111k1897201




                          111k1897201






























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