Triangle Inequality help












0












$begingroup$


Wondering where my logic is going wrong in this assignment:



Show that $||x|-|y|| leq |x-y|$



Using the fact $||x|-|y||, |x-y| geq 0$



It follows $(|x|-|y|)^2 leq (x-y)^2$



Using the fact $|x|^2 = x^2$



$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$



Cancelling down:



$|xy| leq xy$



Which I know is not true. Thanks for any input.






inequality triangle







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
    $endgroup$
    – Crostul
    Dec 17 '18 at 20:32








  • 1




    $begingroup$
    You divided by $-2$ but didn't change the inequality direction.
    $endgroup$
    – orange
    Dec 17 '18 at 20:33
















0












$begingroup$


Wondering where my logic is going wrong in this assignment:



Show that $||x|-|y|| leq |x-y|$



Using the fact $||x|-|y||, |x-y| geq 0$



It follows $(|x|-|y|)^2 leq (x-y)^2$



Using the fact $|x|^2 = x^2$



$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$



Cancelling down:



$|xy| leq xy$



Which I know is not true. Thanks for any input.






inequality triangle







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
    $endgroup$
    – Crostul
    Dec 17 '18 at 20:32








  • 1




    $begingroup$
    You divided by $-2$ but didn't change the inequality direction.
    $endgroup$
    – orange
    Dec 17 '18 at 20:33














0












0








0





$begingroup$


Wondering where my logic is going wrong in this assignment:



Show that $||x|-|y|| leq |x-y|$



Using the fact $||x|-|y||, |x-y| geq 0$



It follows $(|x|-|y|)^2 leq (x-y)^2$



Using the fact $|x|^2 = x^2$



$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$



Cancelling down:



$|xy| leq xy$



Which I know is not true. Thanks for any input.






inequality triangle







share|cite|improve this question









$endgroup$




Wondering where my logic is going wrong in this assignment:



Show that $||x|-|y|| leq |x-y|$



Using the fact $||x|-|y||, |x-y| geq 0$



It follows $(|x|-|y|)^2 leq (x-y)^2$



Using the fact $|x|^2 = x^2$



$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$



Cancelling down:



$|xy| leq xy$



Which I know is not true. Thanks for any input.






inequality triangle




inequality triangle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 20:30









PolynomialCPolynomialC

876




876








  • 2




    $begingroup$
    You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
    $endgroup$
    – Crostul
    Dec 17 '18 at 20:32








  • 1




    $begingroup$
    You divided by $-2$ but didn't change the inequality direction.
    $endgroup$
    – orange
    Dec 17 '18 at 20:33














  • 2




    $begingroup$
    You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
    $endgroup$
    – Crostul
    Dec 17 '18 at 20:32








  • 1




    $begingroup$
    You divided by $-2$ but didn't change the inequality direction.
    $endgroup$
    – orange
    Dec 17 '18 at 20:33








2




2




$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32






$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32






1




1




$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33




$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33










3 Answers
3






active

oldest

votes


















3












$begingroup$

Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thankyou so much!
    $endgroup$
    – PolynomialC
    Dec 17 '18 at 20:33



















1












$begingroup$

You divided by $-2$ but did not change the sign. Another approach.



Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$



Simillarly we have $|y|-|x|leq|x-y|$.



Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.



    However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
    $-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.



    There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Since
      $$ xyle|x||y|$$
      then you have
      $$ -2|x||y|le -2xy. $$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thankyou so much!
        $endgroup$
        – PolynomialC
        Dec 17 '18 at 20:33
















      3












      $begingroup$

      Since
      $$ xyle|x||y|$$
      then you have
      $$ -2|x||y|le -2xy. $$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thankyou so much!
        $endgroup$
        – PolynomialC
        Dec 17 '18 at 20:33














      3












      3








      3





      $begingroup$

      Since
      $$ xyle|x||y|$$
      then you have
      $$ -2|x||y|le -2xy. $$






      share|cite|improve this answer









      $endgroup$



      Since
      $$ xyle|x||y|$$
      then you have
      $$ -2|x||y|le -2xy. $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 17 '18 at 20:32









      xpaulxpaul

      23.3k24655




      23.3k24655












      • $begingroup$
        Thankyou so much!
        $endgroup$
        – PolynomialC
        Dec 17 '18 at 20:33


















      • $begingroup$
        Thankyou so much!
        $endgroup$
        – PolynomialC
        Dec 17 '18 at 20:33
















      $begingroup$
      Thankyou so much!
      $endgroup$
      – PolynomialC
      Dec 17 '18 at 20:33




      $begingroup$
      Thankyou so much!
      $endgroup$
      – PolynomialC
      Dec 17 '18 at 20:33











      1












      $begingroup$

      You divided by $-2$ but did not change the sign. Another approach.



      Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$



      Simillarly we have $|y|-|x|leq|x-y|$.



      Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You divided by $-2$ but did not change the sign. Another approach.



        Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$



        Simillarly we have $|y|-|x|leq|x-y|$.



        Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You divided by $-2$ but did not change the sign. Another approach.



          Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$



          Simillarly we have $|y|-|x|leq|x-y|$.



          Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$






          share|cite|improve this answer









          $endgroup$



          You divided by $-2$ but did not change the sign. Another approach.



          Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$



          Simillarly we have $|y|-|x|leq|x-y|$.



          Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 20:34









          Yadati KiranYadati Kiran

          2,0331621




          2,0331621























              0












              $begingroup$

              First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.



              However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
              $-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.



              There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.



                However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
                $-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.



                There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.



                  However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
                  $-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.



                  There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.






                  share|cite|improve this answer









                  $endgroup$



                  First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.



                  However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
                  $-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.



                  There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 20:50









                  Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena

                  363




                  363






























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