Find the smallest value of the function












3












$begingroup$



There's a function defined as:
$$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
In interval $$Big(0,frac{pi}{2}Big)$$
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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$endgroup$

















    3












    $begingroup$



    There's a function defined as:
    $$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
    In interval $$Big(0,frac{pi}{2}Big)$$
    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.







    $endgroup$















      3












      3








      3





      $begingroup$



      There's a function defined as:
      $$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
      In interval $$Big(0,frac{pi}{2}Big)$$
      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:
      $$f(x) := bigg({1over9}+{32over sin(x)}bigg)bigg({1over32}+{9over cos(x)}bigg).$$
      In interval $$Big(0,frac{pi}{2}Big)$$
      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 3 hours ago









      greedoid

      44.4k1156110




      44.4k1156110






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      asked 4 hours ago









      a_man_with_no_namea_man_with_no_name

      283




      283




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          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          $$1+frac{288}{sin x}$$ 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









          $endgroup$





















            3












            $begingroup$

            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











            $endgroup$





















              0












              $begingroup$

              You can end your idea.



              Indeed, let $sin{x}+cos{x}=t$.



              Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
              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:
              $$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ 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









                3












                $begingroup$

                $$1+frac{288}{sin x}$$ 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









                $endgroup$


















                  3












                  $begingroup$

                  $$1+frac{288}{sin x}$$ 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









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $$1+frac{288}{sin x}$$ 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 4 hours ago









                    Yves DaoustYves Daoust

                    129k675227




                    129k675227























                        3












                        $begingroup$

                        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











                        $endgroup$


















                          3












                          $begingroup$

                          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











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            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 4 hours ago

























                            answered 4 hours ago









                            greedoidgreedoid

                            44.4k1156110




                            44.4k1156110























                                0












                                $begingroup$

                                You can end your idea.



                                Indeed, let $sin{x}+cos{x}=t$.



                                Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                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:
                                $$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  You can end your idea.



                                  Indeed, let $sin{x}+cos{x}=t$.



                                  Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                  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:
                                  $$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    You can end your idea.



                                    Indeed, let $sin{x}+cos{x}=t$.



                                    Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                    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:
                                    $$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!






                                    share|cite|improve this answer









                                    $endgroup$



                                    You can end your idea.



                                    Indeed, let $sin{x}+cos{x}=t$.



                                    Thus, by C-S $$1<t=sin{x}+cos{x}leqsqrt{(1^2+1^2)(sin^2x+cos^2x)}=sqrt2,$$
                                    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:
                                    $$f(x)=frac{1}{288}+frac{2(t+288)}{t^2-1}geqfrac{1}{288}+frac{2(sqrt2+288)}{2-1}=576frac{1}{288}+2sqrt2$$ and we are done!







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 3 hours ago









                                    Michael RozenbergMichael Rozenberg

                                    105k1892198




                                    105k1892198






















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