Proving inequality using double integral

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If it's a known inequality or a duplicate - sorry, iv'e searched the question archive and elsewhere, didn't find anything similar.



Let $f$ be a positive continuous function over the interval $[a,b]$.



Prove the inequality:$$
int_a^bf(x)dx cdot int_a^b frac1f(x)dx geq(b-a)^2
$$



using double integrals.



iv'e tried defining $g(x,y)=fracf(y)f(x)$, and then integrating over the square $[a,b]times[a,b]$, so we get $int_a^bint_a^bfracf(y)f(x)dxdy=int_a^bf(y)dy cdot int_a^b frac1f(x)dx$, and the double integral is then equal to the volume under $g(x,y)$ and over the above mentioned square. didn't manage to prove that the volume is at least $(b-a)^2$.




calculus integration multivariable-calculus definite-integrals




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  • 2




    Cauchy-Schwarz?
    – Lord Shark the Unknown
    2 days ago














up vote
4
down vote

favorite
1












If it's a known inequality or a duplicate - sorry, iv'e searched the question archive and elsewhere, didn't find anything similar.



Let $f$ be a positive continuous function over the interval $[a,b]$.



Prove the inequality:$$
int_a^bf(x)dx cdot int_a^b frac1f(x)dx geq(b-a)^2
$$



using double integrals.



iv'e tried defining $g(x,y)=fracf(y)f(x)$, and then integrating over the square $[a,b]times[a,b]$, so we get $int_a^bint_a^bfracf(y)f(x)dxdy=int_a^bf(y)dy cdot int_a^b frac1f(x)dx$, and the double integral is then equal to the volume under $g(x,y)$ and over the above mentioned square. didn't manage to prove that the volume is at least $(b-a)^2$.




calculus integration multivariable-calculus definite-integrals




share|cite|improve this question















  • 2




    Cauchy-Schwarz?
    – Lord Shark the Unknown
    2 days ago












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





If it's a known inequality or a duplicate - sorry, iv'e searched the question archive and elsewhere, didn't find anything similar.



Let $f$ be a positive continuous function over the interval $[a,b]$.



Prove the inequality:$$
int_a^bf(x)dx cdot int_a^b frac1f(x)dx geq(b-a)^2
$$



using double integrals.



iv'e tried defining $g(x,y)=fracf(y)f(x)$, and then integrating over the square $[a,b]times[a,b]$, so we get $int_a^bint_a^bfracf(y)f(x)dxdy=int_a^bf(y)dy cdot int_a^b frac1f(x)dx$, and the double integral is then equal to the volume under $g(x,y)$ and over the above mentioned square. didn't manage to prove that the volume is at least $(b-a)^2$.




calculus integration multivariable-calculus definite-integrals




share|cite|improve this question











If it's a known inequality or a duplicate - sorry, iv'e searched the question archive and elsewhere, didn't find anything similar.



Let $f$ be a positive continuous function over the interval $[a,b]$.



Prove the inequality:$$
int_a^bf(x)dx cdot int_a^b frac1f(x)dx geq(b-a)^2
$$



using double integrals.



iv'e tried defining $g(x,y)=fracf(y)f(x)$, and then integrating over the square $[a,b]times[a,b]$, so we get $int_a^bint_a^bfracf(y)f(x)dxdy=int_a^bf(y)dy cdot int_a^b frac1f(x)dx$, and the double integral is then equal to the volume under $g(x,y)$ and over the above mentioned square. didn't manage to prove that the volume is at least $(b-a)^2$.





calculus integration multivariable-calculus definite-integrals





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asked 2 days ago









juleand

896




896







  • 2




    Cauchy-Schwarz?
    – Lord Shark the Unknown
    2 days ago












  • 2




    Cauchy-Schwarz?
    – Lord Shark the Unknown
    2 days ago







2




2




Cauchy-Schwarz?
– Lord Shark the Unknown
2 days ago




Cauchy-Schwarz?
– Lord Shark the Unknown
2 days ago










2 Answers
2






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up vote
3
down vote



accepted










The double integral method:



Proof. $blacktriangleleft$
beginalign*
int_a^b f int_a^b frac 1 f
&= int_a^b f(x)mathrm dx int_a^b frac 1f(y) mathrm dy\
&= iint_[a,b]^2 frac f(x)f(y) mathrm dx mathrm dy.
endalign*
Also,
beginalign*
int_a^b f int_a^b frac 1 f
& = int_a^b f(y)mathrm dy int_a^b frac 1f(x)mathrm dx\
&= iint_[a,b]^2 frac f(y)f(x) mathrm dx mathrm dy
endalign*
Hence
beginalign*
int_a^b f int_a^b frac 1 f
&= frac12 iint_[a,b]^2 left( frac f(x) f(y) + frac f(y) f(x)right)mathrm dx mathrm dy\
&geqslant iint_[a,b]^2 sqrtfrac f(x) f(y) cdot frac f(y) f(x) mathrm dx mathrm dy \
&= (b-a)^2. blacktriangleright
endalign*






share|cite|improve this answer























  • Thank you very much. could you please explain the inequality in the row before the last one?
    – juleand
    2 days ago











  • @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
    – xbh
    2 days ago










  • got it now, thanks!
    – juleand
    2 days ago

















up vote
0
down vote













Using Cauchy-Schwarz as suggested by Lord Shark the Unknown,



beginmultlineleft|b-aright|^2=left|int_a^bdxright|^2=left|int_a^bsqrtf(x)frac1sqrtf(x)dxright|^2leqint_a^bleft|sqrtf(x)right|^2dxint_a^bleft|frac1sqrtf(x)right|^2dx\=int_a^bf(x)dxint_a^bfrac1f(x)dx.endmultline



We have used the fact that the function $f$ is positive, but we have not made use of the fact that it is continuous.
Indeed, this should work for any integrable positive function $f$.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    The double integral method:



    Proof. $blacktriangleleft$
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= int_a^b f(x)mathrm dx int_a^b frac 1f(y) mathrm dy\
    &= iint_[a,b]^2 frac f(x)f(y) mathrm dx mathrm dy.
    endalign*
    Also,
    beginalign*
    int_a^b f int_a^b frac 1 f
    & = int_a^b f(y)mathrm dy int_a^b frac 1f(x)mathrm dx\
    &= iint_[a,b]^2 frac f(y)f(x) mathrm dx mathrm dy
    endalign*
    Hence
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= frac12 iint_[a,b]^2 left( frac f(x) f(y) + frac f(y) f(x)right)mathrm dx mathrm dy\
    &geqslant iint_[a,b]^2 sqrtfrac f(x) f(y) cdot frac f(y) f(x) mathrm dx mathrm dy \
    &= (b-a)^2. blacktriangleright
    endalign*






    share|cite|improve this answer























    • Thank you very much. could you please explain the inequality in the row before the last one?
      – juleand
      2 days ago











    • @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
      – xbh
      2 days ago










    • got it now, thanks!
      – juleand
      2 days ago














    up vote
    3
    down vote



    accepted










    The double integral method:



    Proof. $blacktriangleleft$
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= int_a^b f(x)mathrm dx int_a^b frac 1f(y) mathrm dy\
    &= iint_[a,b]^2 frac f(x)f(y) mathrm dx mathrm dy.
    endalign*
    Also,
    beginalign*
    int_a^b f int_a^b frac 1 f
    & = int_a^b f(y)mathrm dy int_a^b frac 1f(x)mathrm dx\
    &= iint_[a,b]^2 frac f(y)f(x) mathrm dx mathrm dy
    endalign*
    Hence
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= frac12 iint_[a,b]^2 left( frac f(x) f(y) + frac f(y) f(x)right)mathrm dx mathrm dy\
    &geqslant iint_[a,b]^2 sqrtfrac f(x) f(y) cdot frac f(y) f(x) mathrm dx mathrm dy \
    &= (b-a)^2. blacktriangleright
    endalign*






    share|cite|improve this answer























    • Thank you very much. could you please explain the inequality in the row before the last one?
      – juleand
      2 days ago











    • @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
      – xbh
      2 days ago










    • got it now, thanks!
      – juleand
      2 days ago












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    The double integral method:



    Proof. $blacktriangleleft$
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= int_a^b f(x)mathrm dx int_a^b frac 1f(y) mathrm dy\
    &= iint_[a,b]^2 frac f(x)f(y) mathrm dx mathrm dy.
    endalign*
    Also,
    beginalign*
    int_a^b f int_a^b frac 1 f
    & = int_a^b f(y)mathrm dy int_a^b frac 1f(x)mathrm dx\
    &= iint_[a,b]^2 frac f(y)f(x) mathrm dx mathrm dy
    endalign*
    Hence
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= frac12 iint_[a,b]^2 left( frac f(x) f(y) + frac f(y) f(x)right)mathrm dx mathrm dy\
    &geqslant iint_[a,b]^2 sqrtfrac f(x) f(y) cdot frac f(y) f(x) mathrm dx mathrm dy \
    &= (b-a)^2. blacktriangleright
    endalign*






    share|cite|improve this answer















    The double integral method:



    Proof. $blacktriangleleft$
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= int_a^b f(x)mathrm dx int_a^b frac 1f(y) mathrm dy\
    &= iint_[a,b]^2 frac f(x)f(y) mathrm dx mathrm dy.
    endalign*
    Also,
    beginalign*
    int_a^b f int_a^b frac 1 f
    & = int_a^b f(y)mathrm dy int_a^b frac 1f(x)mathrm dx\
    &= iint_[a,b]^2 frac f(y)f(x) mathrm dx mathrm dy
    endalign*
    Hence
    beginalign*
    int_a^b f int_a^b frac 1 f
    &= frac12 iint_[a,b]^2 left( frac f(x) f(y) + frac f(y) f(x)right)mathrm dx mathrm dy\
    &geqslant iint_[a,b]^2 sqrtfrac f(x) f(y) cdot frac f(y) f(x) mathrm dx mathrm dy \
    &= (b-a)^2. blacktriangleright
    endalign*







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago


























    answered 2 days ago









    xbh

    9156




    9156











    • Thank you very much. could you please explain the inequality in the row before the last one?
      – juleand
      2 days ago











    • @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
      – xbh
      2 days ago










    • got it now, thanks!
      – juleand
      2 days ago
















    • Thank you very much. could you please explain the inequality in the row before the last one?
      – juleand
      2 days ago











    • @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
      – xbh
      2 days ago










    • got it now, thanks!
      – juleand
      2 days ago















    Thank you very much. could you please explain the inequality in the row before the last one?
    – juleand
    2 days ago





    Thank you very much. could you please explain the inequality in the row before the last one?
    – juleand
    2 days ago













    @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
    – xbh
    2 days ago




    @juleand My pleasure. I've used AM-GM inequality. It's applicable because $f$ is positive on $[a,b]$
    – xbh
    2 days ago












    got it now, thanks!
    – juleand
    2 days ago




    got it now, thanks!
    – juleand
    2 days ago










    up vote
    0
    down vote













    Using Cauchy-Schwarz as suggested by Lord Shark the Unknown,



    beginmultlineleft|b-aright|^2=left|int_a^bdxright|^2=left|int_a^bsqrtf(x)frac1sqrtf(x)dxright|^2leqint_a^bleft|sqrtf(x)right|^2dxint_a^bleft|frac1sqrtf(x)right|^2dx\=int_a^bf(x)dxint_a^bfrac1f(x)dx.endmultline



    We have used the fact that the function $f$ is positive, but we have not made use of the fact that it is continuous.
    Indeed, this should work for any integrable positive function $f$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Using Cauchy-Schwarz as suggested by Lord Shark the Unknown,



      beginmultlineleft|b-aright|^2=left|int_a^bdxright|^2=left|int_a^bsqrtf(x)frac1sqrtf(x)dxright|^2leqint_a^bleft|sqrtf(x)right|^2dxint_a^bleft|frac1sqrtf(x)right|^2dx\=int_a^bf(x)dxint_a^bfrac1f(x)dx.endmultline



      We have used the fact that the function $f$ is positive, but we have not made use of the fact that it is continuous.
      Indeed, this should work for any integrable positive function $f$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Using Cauchy-Schwarz as suggested by Lord Shark the Unknown,



        beginmultlineleft|b-aright|^2=left|int_a^bdxright|^2=left|int_a^bsqrtf(x)frac1sqrtf(x)dxright|^2leqint_a^bleft|sqrtf(x)right|^2dxint_a^bleft|frac1sqrtf(x)right|^2dx\=int_a^bf(x)dxint_a^bfrac1f(x)dx.endmultline



        We have used the fact that the function $f$ is positive, but we have not made use of the fact that it is continuous.
        Indeed, this should work for any integrable positive function $f$.






        share|cite|improve this answer













        Using Cauchy-Schwarz as suggested by Lord Shark the Unknown,



        beginmultlineleft|b-aright|^2=left|int_a^bdxright|^2=left|int_a^bsqrtf(x)frac1sqrtf(x)dxright|^2leqint_a^bleft|sqrtf(x)right|^2dxint_a^bleft|frac1sqrtf(x)right|^2dx\=int_a^bf(x)dxint_a^bfrac1f(x)dx.endmultline



        We have used the fact that the function $f$ is positive, but we have not made use of the fact that it is continuous.
        Indeed, this should work for any integrable positive function $f$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        parsiad

        15.9k32153




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