Find the smallest value of $f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$...












4












$begingroup$



There's a function defined as:
$$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
In interval $$left(0,frac{pi}{2}right)$$
Find the smallest value (Save only its integer value)




I've managed to come to this
$${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
How can I find the smallest value now?






trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals







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    4












    $begingroup$



    There's a function defined as:
    $$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
    In interval $$left(0,frac{pi}{2}right)$$
    Find the smallest value (Save only its integer value)




    I've managed to come to this
    $${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
    How can I find the smallest value now?






    trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals







    share|cite|improve this question









    New contributor




    a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$



      There's a function defined as:
      $$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
      In interval $$left(0,frac{pi}{2}right)$$
      Find the smallest value (Save only its integer value)




      I've managed to come to this
      $${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
      How can I find the smallest value now?






      trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals







      share|cite|improve this question









      New contributor




      a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      There's a function defined as:
      $$f(x) := left({1over9}+{32over sin(x)}right)left({1over32}+{9over cos(x)}right)$$
      In interval $$left(0,frac{pi}{2}right)$$
      Find the smallest value (Save only its integer value)




      I've managed to come to this
      $${1over288}+{{2(sin(x)+cos(x))+576}over(sin(x)+cos(x))^2-1}$$
      How can I find the smallest value now?






      trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals




      trigonometry inequality optimization cauchy-schwarz-inequality trigonometric-integrals






      share|cite|improve this question









      New contributor




      a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Blue

      48.6k870154




      48.6k870154






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      asked 7 hours ago









      a_man_with_no_namea_man_with_no_name

      333




      333




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          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:



          $$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$



          $$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$



          We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
          with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$






          share|cite|improve this answer











          $endgroup$





















            5












            $begingroup$

            $$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric



            $$1+frac{288}{cos x}$$ is increasing.



            Hence the minimum occurs ar $x=dfracpi4$.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Here's how to continue with your idea:
              Let $sin{x}+cos{x}=t$.
              Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
              where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
              $$
              f(x)
              =frac{1}{288}+frac{2(t+288)}{t^2-1}
              ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
              =576 + frac{1}{288} + 2sqrt{2}$$
              and we are done!






              share|cite|improve this answer











              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:



                $$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$



                $$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$



                We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
                with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$






                share|cite|improve this answer











                $endgroup$


















                  6












                  $begingroup$

                  Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:



                  $$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$



                  $$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$



                  We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
                  with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$






                  share|cite|improve this answer











                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:



                    $$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$



                    $$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$



                    We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
                    with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$






                    share|cite|improve this answer











                    $endgroup$



                    Since both $cos$ and $sin $ are positive in $(o,{piover 2})$ we can use Cauchy inequaliy:



                    $$ (a^2+b^2)(c^2+d^2)geq (ac+bd)^2$$



                    $$bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg)geq bigg({1over sqrt{288}}+{sqrt{288}over sqrt{sin(x)cos(x)}}bigg)^2geq bigg({1over 12sqrt{2}}+24bigg)^2$$



                    We used here $$sin(x)cos(x)= {1over 2}sin (2x) leq {1over 2}$$
                    with equality at $x={pi over 4}$. So $$y_{min} = bigg({1over 12sqrt{2}}+24bigg)^2$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 6 hours ago

























                    answered 7 hours ago









                    greedoidgreedoid

                    44.4k1156110




                    44.4k1156110























                        5












                        $begingroup$

                        $$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric



                        $$1+frac{288}{cos x}$$ is increasing.



                        Hence the minimum occurs ar $x=dfracpi4$.






                        share|cite|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          $$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric



                          $$1+frac{288}{cos x}$$ is increasing.



                          Hence the minimum occurs ar $x=dfracpi4$.






                          share|cite|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            $$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric



                            $$1+frac{288}{cos x}$$ is increasing.



                            Hence the minimum occurs ar $x=dfracpi4$.






                            share|cite|improve this answer









                            $endgroup$



                            $$1+frac{288}{sin x}$$ is a decreasing function in $left(0,dfracpi2right)$ and its symmetric



                            $$1+frac{288}{cos x}$$ is increasing.



                            Hence the minimum occurs ar $x=dfracpi4$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            Yves DaoustYves Daoust

                            129k675227




                            129k675227























                                1












                                $begingroup$

                                Here's how to continue with your idea:
                                Let $sin{x}+cos{x}=t$.
                                Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
                                $$
                                f(x)
                                =frac{1}{288}+frac{2(t+288)}{t^2-1}
                                ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
                                =576 + frac{1}{288} + 2sqrt{2}$$
                                and we are done!






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  Here's how to continue with your idea:
                                  Let $sin{x}+cos{x}=t$.
                                  Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                  where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
                                  $$
                                  f(x)
                                  =frac{1}{288}+frac{2(t+288)}{t^2-1}
                                  ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
                                  =576 + frac{1}{288} + 2sqrt{2}$$
                                  and we are done!






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Here's how to continue with your idea:
                                    Let $sin{x}+cos{x}=t$.
                                    Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                    where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
                                    $$
                                    f(x)
                                    =frac{1}{288}+frac{2(t+288)}{t^2-1}
                                    ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
                                    =576 + frac{1}{288} + 2sqrt{2}$$
                                    and we are done!






                                    share|cite|improve this answer











                                    $endgroup$



                                    Here's how to continue with your idea:
                                    Let $sin{x}+cos{x}=t$.
                                    Then, by the Cauchy-Schwartz inequality we have $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                    where the equality occurs for $x=frac{pi}{4},$ and since $$left(frac{t+288}{t^2-1}right)'=-frac{x^2+576x+1}{(x^2-1)^2}<0,$$ we obtain
                                    $$
                                    f(x)
                                    =frac{1}{288}+frac{2(t+288)}{t^2-1}
                                    ge frac{1}{288}+frac{2(sqrt{2}+288)}{2-1}
                                    =576 + frac{1}{288} + 2sqrt{2}$$
                                    and we are done!







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 1 hour ago









                                    Viktor Glombik

                                    9911527




                                    9911527










                                    answered 5 hours ago









                                    Michael RozenbergMichael Rozenberg

                                    105k1892198




                                    105k1892198






















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