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Solve the given inequality below in the body.


Rational/Quadratic InequalityProof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$Is $int_x^inftye^-fract^22 < frac1xe^-fracx^22$?Is the inequality solution below legal?Young's inequality or something similarTypical Absolute value inequalityWhich is the proper way to solve the inequality problems?Solve the inequality $sin x > ln x$






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$



$$frac1-3 le frac 12$$




Let’s consider $|x|=y$



So $$frac1y-3-frac 12 le 0$$



$$frac2-y+3y-3 le 0$$
$$fracy-5y-3 ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!










share|cite|improve this question









$endgroup$




















    4












    $begingroup$



    $$frac1-3 le frac 12$$




    Let’s consider $|x|=y$



    So $$frac1y-3-frac 12 le 0$$



    $$frac2-y+3y-3 le 0$$
    $$fracy-5y-3 ge 0$$
    $$y in (-infty , 3)cup [5, infty)$$
    Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!










    share|cite|improve this question









    $endgroup$
















      4












      4








      4


      1



      $begingroup$



      $$frac1-3 le frac 12$$




      Let’s consider $|x|=y$



      So $$frac1y-3-frac 12 le 0$$



      $$frac2-y+3y-3 le 0$$
      $$fracy-5y-3 ge 0$$
      $$y in (-infty , 3)cup [5, infty)$$
      Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!










      share|cite|improve this question









      $endgroup$





      $$frac1-3 le frac 12$$




      Let’s consider $|x|=y$



      So $$frac1y-3-frac 12 le 0$$



      $$frac2-y+3y-3 le 0$$
      $$fracy-5y-3 ge 0$$
      $$y in (-infty , 3)cup [5, infty)$$
      Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!







      inequality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 8 hours ago









      Aditya Aditya

      39410 bronze badges




      39410 bronze badges























          3 Answers
          3






          active

          oldest

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          4














          $begingroup$

          Anyway, in your case, $yge 0$, so you should transform all this to
          $$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
          or, as a set, the union of three intervals:
          $$(-infty,-5]cup(-3,3)cup[5,+infty). $$






          share|cite|improve this answer









          $endgroup$






















            2














            $begingroup$

            The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$



            are a good place to start from.






            share|cite|improve this answer











            $endgroup$














            • $begingroup$
              In your first equivalence, the inequalities should be strict.
              $endgroup$
              – Bernard
              8 hours ago










            • $begingroup$
              @Bernard Yes, thank you.
              $endgroup$
              – Gae. S.
              8 hours ago










            • $begingroup$
              Even $B$ should be positive, since $|A|ge 0$.
              $endgroup$
              – Bernard
              8 hours ago










            • $begingroup$
              @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
              $endgroup$
              – Gae. S.
              8 hours ago










            • $begingroup$
              There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
              $endgroup$
              – Bernard
              8 hours ago


















            1














            $begingroup$

            For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation



            $$frac1-3=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.



            So the solution set is



            $$(-3,3)cup(-infty,-5]cup[5,infty).$$






            share|cite|improve this answer











            $endgroup$

















              Your Answer








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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4














              $begingroup$

              Anyway, in your case, $yge 0$, so you should transform all this to
              $$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
              or, as a set, the union of three intervals:
              $$(-infty,-5]cup(-3,3)cup[5,+infty). $$






              share|cite|improve this answer









              $endgroup$



















                4














                $begingroup$

                Anyway, in your case, $yge 0$, so you should transform all this to
                $$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
                or, as a set, the union of three intervals:
                $$(-infty,-5]cup(-3,3)cup[5,+infty). $$






                share|cite|improve this answer









                $endgroup$

















                  4














                  4










                  4







                  $begingroup$

                  Anyway, in your case, $yge 0$, so you should transform all this to
                  $$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
                  or, as a set, the union of three intervals:
                  $$(-infty,-5]cup(-3,3)cup[5,+infty). $$






                  share|cite|improve this answer









                  $endgroup$



                  Anyway, in your case, $yge 0$, so you should transform all this to
                  $$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
                  or, as a set, the union of three intervals:
                  $$(-infty,-5]cup(-3,3)cup[5,+infty). $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  BernardBernard

                  132k7 gold badges43 silver badges126 bronze badges




                  132k7 gold badges43 silver badges126 bronze badges


























                      2














                      $begingroup$

                      The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$



                      are a good place to start from.






                      share|cite|improve this answer











                      $endgroup$














                      • $begingroup$
                        In your first equivalence, the inequalities should be strict.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard Yes, thank you.
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        Even $B$ should be positive, since $|A|ge 0$.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                        $endgroup$
                        – Bernard
                        8 hours ago















                      2














                      $begingroup$

                      The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$



                      are a good place to start from.






                      share|cite|improve this answer











                      $endgroup$














                      • $begingroup$
                        In your first equivalence, the inequalities should be strict.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard Yes, thank you.
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        Even $B$ should be positive, since $|A|ge 0$.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                        $endgroup$
                        – Bernard
                        8 hours ago













                      2














                      2










                      2







                      $begingroup$

                      The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$



                      are a good place to start from.






                      share|cite|improve this answer











                      $endgroup$



                      The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$



                      are a good place to start from.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 8 hours ago

























                      answered 8 hours ago









                      Gae. S.Gae. S.

                      2,0648 silver badges17 bronze badges




                      2,0648 silver badges17 bronze badges














                      • $begingroup$
                        In your first equivalence, the inequalities should be strict.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard Yes, thank you.
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        Even $B$ should be positive, since $|A|ge 0$.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                        $endgroup$
                        – Bernard
                        8 hours ago
















                      • $begingroup$
                        In your first equivalence, the inequalities should be strict.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard Yes, thank you.
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        Even $B$ should be positive, since $|A|ge 0$.
                        $endgroup$
                        – Bernard
                        8 hours ago










                      • $begingroup$
                        @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                        $endgroup$
                        – Gae. S.
                        8 hours ago










                      • $begingroup$
                        There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                        $endgroup$
                        – Bernard
                        8 hours ago















                      $begingroup$
                      In your first equivalence, the inequalities should be strict.
                      $endgroup$
                      – Bernard
                      8 hours ago




                      $begingroup$
                      In your first equivalence, the inequalities should be strict.
                      $endgroup$
                      – Bernard
                      8 hours ago












                      $begingroup$
                      @Bernard Yes, thank you.
                      $endgroup$
                      – Gae. S.
                      8 hours ago




                      $begingroup$
                      @Bernard Yes, thank you.
                      $endgroup$
                      – Gae. S.
                      8 hours ago












                      $begingroup$
                      Even $B$ should be positive, since $|A|ge 0$.
                      $endgroup$
                      – Bernard
                      8 hours ago




                      $begingroup$
                      Even $B$ should be positive, since $|A|ge 0$.
                      $endgroup$
                      – Bernard
                      8 hours ago












                      $begingroup$
                      @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                      $endgroup$
                      – Gae. S.
                      8 hours ago




                      $begingroup$
                      @Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
                      $endgroup$
                      – Gae. S.
                      8 hours ago












                      $begingroup$
                      There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                      $endgroup$
                      – Bernard
                      8 hours ago




                      $begingroup$
                      There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
                      $endgroup$
                      – Bernard
                      8 hours ago











                      1














                      $begingroup$

                      For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation



                      $$frac1-3=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.



                      So the solution set is



                      $$(-3,3)cup(-infty,-5]cup[5,infty).$$






                      share|cite|improve this answer











                      $endgroup$



















                        1














                        $begingroup$

                        For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation



                        $$frac1-3=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.



                        So the solution set is



                        $$(-3,3)cup(-infty,-5]cup[5,infty).$$






                        share|cite|improve this answer











                        $endgroup$

















                          1














                          1










                          1







                          $begingroup$

                          For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation



                          $$frac1-3=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.



                          So the solution set is



                          $$(-3,3)cup(-infty,-5]cup[5,infty).$$






                          share|cite|improve this answer











                          $endgroup$



                          For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation



                          $$frac1-3=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.



                          So the solution set is



                          $$(-3,3)cup(-infty,-5]cup[5,infty).$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 6 hours ago

























                          answered 8 hours ago









                          Yves DaoustYves Daoust

                          145k10 gold badges89 silver badges248 bronze badges




                          145k10 gold badges89 silver badges248 bronze badges






























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