Inequality similar to Minkowski












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Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
The form is very close to Minkowski's inequality,but I can't really show this.






algebra-precalculus inequality







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    $begingroup$


    Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
    The form is very close to Minkowski's inequality,but I can't really show this.






    algebra-precalculus inequality







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      0





      $begingroup$


      Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
      The form is very close to Minkowski's inequality,but I can't really show this.






      algebra-precalculus inequality







      share|cite|improve this question









      $endgroup$




      Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
      The form is very close to Minkowski's inequality,but I can't really show this.






      algebra-precalculus inequality




      algebra-precalculus inequality






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 at 6:03









      Legend KillerLegend Killer

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          $begingroup$

          First of all, you need $pgeq1$.



          If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and



          $$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$






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            $begingroup$

            First of all, you need $pgeq1$.



            If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and



            $$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              First of all, you need $pgeq1$.



              If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and



              $$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                First of all, you need $pgeq1$.



                If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and



                $$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$






                share|cite|improve this answer











                $endgroup$



                First of all, you need $pgeq1$.



                If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and



                $$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 21 at 9:53

























                answered Jan 21 at 6:15









                Michael RozenbergMichael Rozenberg

                107k1894198




                107k1894198






























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