If $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove than $int_0^1(f(x))^2dxgeqfrac43$Prove that $int_0^1|f(x)f'(x)|dxleqfrac12int_0^1|f'(x)|^2dx$Integral inequality $int_0^1h(x)^2dxint_0^1x^2h(x)^2dxge cleft(int_0^1h(x)dxright)^4$Prove that $int_0^1f(x)^2dxgeq 4$$int_0^1f(x)cdot g(x)dx=int_0^1f(x)dxcdot int_0^1g(x)dx$Prove that:$f(f(x)) = x^2 implies int_0^1(f(x))^2dx geq frac313$how can we explain that $int_0^L1 + frac12(u'(x))^2dx - L = int_0^Lfrac12(u'(x))^2dx$?For a continuous function with $int_0^1 f(x)dx=1$, and $M=maxf(x)$, show that $1-frac12Mgeqint_0^1xf(x)dxgeqfrac12M$Let $f:mathbbRrightarrowmathbbR$ be a continuous function such that $int_x^1f(t)dtgeq(1-x)^2$. Prove that $f(1)=0$Prove that $int_0^1f(x)arctan x dx=fracπ8int_0^1f(x)dx$Prove that $f(x)=0$ for any $xin[0,1]$.

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If $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove than $int_0^1(f(x))^2dxgeqfrac43$


Prove that $int_0^1|f(x)f'(x)|dxleqfrac12int_0^1|f'(x)|^2dx$Integral inequality $int_0^1h(x)^2dxint_0^1x^2h(x)^2dxge cleft(int_0^1h(x)dxright)^4$Prove that $int_0^1f(x)^2dxgeq 4$$int_0^1f(x)cdot g(x)dx=int_0^1f(x)dxcdot int_0^1g(x)dx$Prove that:$f(f(x)) = x^2 implies int_0^1(f(x))^2dx geq frac313$how can we explain that $int_0^L1 + frac12(u'(x))^2dx - L = int_0^Lfrac12(u'(x))^2dx$?For a continuous function with $int_0^1 f(x)dx=1$, and $M=maxf(x)$, show that $1-frac12Mgeqint_0^1xf(x)dxgeqfrac12M$Let $f:mathbbRrightarrowmathbbR$ be a continuous function such that $int_x^1f(t)dtgeq(1-x)^2$. Prove that $f(1)=0$Prove that $int_0^1f(x)arctan x dx=fracπ8int_0^1f(x)dx$Prove that $f(x)=0$ for any $xin[0,1]$.













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$begingroup$



If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.




However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$



    If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.




    However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$



      If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.




      However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?










      share|cite|improve this question











      $endgroup$





      If $f:[0,1]rightarrowmathbbR$ is a continuous function such that $2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$, prove that $int_0^1(f(x))^2dxgeqfrac43$.




      However, using CBS I proved that $int_0^1(f(x))^2dxint_0^1x^2dxgeq(int_0^1(f(x))xdx)^2geqfrac14(int_0^1(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$frac43geqint_0^1(f(x))^2dx$. Is there a mistake in the task or in my proof?







      integration analysis inequality definite-integrals continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Michael Rozenberg

      107k1895200




      107k1895200










      asked Mar 7 at 8:42









      Septimiu CristianSeptimiu Cristian

      1




      1




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
          $$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
          but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.



          On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
          $$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
          which implies
          $$int_0^1(f(x))^2dxleq frac43.$$






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.



            Otherwise, your solution is not full.



            Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
            and since $intlimits_0^1x^2dx=frac13$, we obtain:
            $$intlimits_0^1f(x)^2dxleqfrac43.$$






            share|cite|improve this answer











            $endgroup$












              Your Answer





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              2 Answers
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              2 Answers
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              active

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              3












              $begingroup$

              Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
              $$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
              but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.



              On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
              $$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
              which implies
              $$int_0^1(f(x))^2dxleq frac43.$$






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
                $$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
                but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.



                On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
                $$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
                which implies
                $$int_0^1(f(x))^2dxleq frac43.$$






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
                  $$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
                  but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.



                  On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
                  $$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
                  which implies
                  $$int_0^1(f(x))^2dxleq frac43.$$






                  share|cite|improve this answer











                  $endgroup$



                  Yes, the inequality $int_0^1(f(x))^2dxgeqfrac43$ should be reversed. Note that $f=0$ satisfies
                  $$2int_0^1 xf(x)dxgeqint_0^1(f(x))^2dx$$
                  but $int_0^1(f(x))^2dxgeqfrac43$ does not hold.



                  On the other hand, by Cauchy-Schwarz (note that $int_0^1 2xf(x)dxgeq 0$)
                  $$left(int_0^1(f(x))^2dxright)^2leqleft(int_0^1 2xf(x)dxright)^2leq int_0^14x^2dxint_0^1(f(x))^2dx=frac43int_0^1(f(x))^2dx$$
                  which implies
                  $$int_0^1(f(x))^2dxleq frac43.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 7 at 9:02

























                  answered Mar 7 at 8:47









                  Robert ZRobert Z

                  100k1069140




                  100k1069140





















                      1












                      $begingroup$

                      You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.



                      Otherwise, your solution is not full.



                      Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
                      and since $intlimits_0^1x^2dx=frac13$, we obtain:
                      $$intlimits_0^1f(x)^2dxleqfrac43.$$






                      share|cite|improve this answer











                      $endgroup$

















                        1












                        $begingroup$

                        You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.



                        Otherwise, your solution is not full.



                        Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
                        and since $intlimits_0^1x^2dx=frac13$, we obtain:
                        $$intlimits_0^1f(x)^2dxleqfrac43.$$






                        share|cite|improve this answer











                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.



                          Otherwise, your solution is not full.



                          Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
                          and since $intlimits_0^1x^2dx=frac13$, we obtain:
                          $$intlimits_0^1f(x)^2dxleqfrac43.$$






                          share|cite|improve this answer











                          $endgroup$



                          You need to write that the condition gives $intlimits_0^1xf(x)dxgeq0$.



                          Otherwise, your solution is not full.



                          Now, by C-S $$intlimits_0^1f(x)^2dxintlimits_0^1x^2dxgeqleft(intlimits_0^1xf(x)dxright)^2geqfrac14left(intlimits_0^1f(x)^2dxright)^2$$
                          and since $intlimits_0^1x^2dx=frac13$, we obtain:
                          $$intlimits_0^1f(x)^2dxleqfrac43.$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 days ago

























                          answered Mar 7 at 9:01









                          Michael RozenbergMichael Rozenberg

                          107k1895200




                          107k1895200



























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