Proving an inequality of modulus of complex numbers












2












$begingroup$


Let $x,y,z$ be distinct complex numbers such that $$y=tx+(1-t)z$$ ; where $t in (0,1)$



Prove that $$frac{|z|-|y|}{|z-y|} geq frac{|z|-|x|}{|z-x|} geq frac{|y|-|x|}{|y-x|} $$



I tried this question by finding $t$ in terms of $x,y,z$. But I'm getting an inequality which doesn't lead to the answer.



$$|t|= frac{|y-z|}{|x-z|} $$



So we have $$0 leq|y-z|leq|x-z|$$



Now it is clear that $$0 leq|y|-|z|leq|x-z|$$



After this ? I;m stuck. Any ideas ?






inequality complex-numbers







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$endgroup$

















    2












    $begingroup$


    Let $x,y,z$ be distinct complex numbers such that $$y=tx+(1-t)z$$ ; where $t in (0,1)$



    Prove that $$frac{|z|-|y|}{|z-y|} geq frac{|z|-|x|}{|z-x|} geq frac{|y|-|x|}{|y-x|} $$



    I tried this question by finding $t$ in terms of $x,y,z$. But I'm getting an inequality which doesn't lead to the answer.



    $$|t|= frac{|y-z|}{|x-z|} $$



    So we have $$0 leq|y-z|leq|x-z|$$



    Now it is clear that $$0 leq|y|-|z|leq|x-z|$$



    After this ? I;m stuck. Any ideas ?






    inequality complex-numbers







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $x,y,z$ be distinct complex numbers such that $$y=tx+(1-t)z$$ ; where $t in (0,1)$



      Prove that $$frac{|z|-|y|}{|z-y|} geq frac{|z|-|x|}{|z-x|} geq frac{|y|-|x|}{|y-x|} $$



      I tried this question by finding $t$ in terms of $x,y,z$. But I'm getting an inequality which doesn't lead to the answer.



      $$|t|= frac{|y-z|}{|x-z|} $$



      So we have $$0 leq|y-z|leq|x-z|$$



      Now it is clear that $$0 leq|y|-|z|leq|x-z|$$



      After this ? I;m stuck. Any ideas ?






      inequality complex-numbers







      share|cite|improve this question









      $endgroup$




      Let $x,y,z$ be distinct complex numbers such that $$y=tx+(1-t)z$$ ; where $t in (0,1)$



      Prove that $$frac{|z|-|y|}{|z-y|} geq frac{|z|-|x|}{|z-x|} geq frac{|y|-|x|}{|y-x|} $$



      I tried this question by finding $t$ in terms of $x,y,z$. But I'm getting an inequality which doesn't lead to the answer.



      $$|t|= frac{|y-z|}{|x-z|} $$



      So we have $$0 leq|y-z|leq|x-z|$$



      Now it is clear that $$0 leq|y|-|z|leq|x-z|$$



      After this ? I;m stuck. Any ideas ?






      inequality complex-numbers




      inequality complex-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 28 '18 at 17:23









      Angelo MarkAngelo Mark

      4,04421741




      4,04421741






















          1 Answer
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          $begingroup$

          Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality,
          $$|z|-|y|geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$
          and so we now have
          $$frac{|z|-|y|}{|z-y|}geqfrac{t(|z|-|x|)}{t|z-x|}=frac{|z|-|x|}{|z-x|}.$$
          The remaining inequality should be similar.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality,
            $$|z|-|y|geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$
            and so we now have
            $$frac{|z|-|y|}{|z-y|}geqfrac{t(|z|-|x|)}{t|z-x|}=frac{|z|-|x|}{|z-x|}.$$
            The remaining inequality should be similar.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality,
              $$|z|-|y|geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$
              and so we now have
              $$frac{|z|-|y|}{|z-y|}geqfrac{t(|z|-|x|)}{t|z-x|}=frac{|z|-|x|}{|z-x|}.$$
              The remaining inequality should be similar.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality,
                $$|z|-|y|geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$
                and so we now have
                $$frac{|z|-|y|}{|z-y|}geqfrac{t(|z|-|x|)}{t|z-x|}=frac{|z|-|x|}{|z-x|}.$$
                The remaining inequality should be similar.






                share|cite|improve this answer









                $endgroup$



                Note that $|z-y|=|-tx+tz|=t|z-x|$. Meanwhile, by the triangle inequality,
                $$|z|-|y|geq|z|-t|x|-(1-t)|z|=t(|z|-|x|)$$
                and so we now have
                $$frac{|z|-|y|}{|z-y|}geqfrac{t(|z|-|x|)}{t|z-x|}=frac{|z|-|x|}{|z-x|}.$$
                The remaining inequality should be similar.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 18:29









                Ben WBen W

                2,276615




                2,276615






























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