Prove $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $...












1












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Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.






calculus algebra-precalculus trigonometry inequality contest-math







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closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    is the hypothesis true for every $theta$?
    $endgroup$
    – mookid
    Sep 19 '14 at 15:33
















1












$begingroup$


Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.






calculus algebra-precalculus trigonometry inequality contest-math







share|cite|improve this question











$endgroup$



closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    is the hypothesis true for every $theta$?
    $endgroup$
    – mookid
    Sep 19 '14 at 15:33














1












1








1





$begingroup$


Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.






calculus algebra-precalculus trigonometry inequality contest-math







share|cite|improve this question











$endgroup$




Prove that $mcos^2{theta} + nsin^2{theta} < l implies sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta} < sqrt{l} $ for every $m, n, l >0$.






calculus algebra-precalculus trigonometry inequality contest-math




calculus algebra-precalculus trigonometry inequality contest-math






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 18:47









Martin Sleziak

45.1k10123277




45.1k10123277










asked Sep 19 '14 at 15:30









user62029user62029

570523




570523




closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Did, metamorphy, Adrian Keister, clathratus, mrtaurho Jan 19 at 21:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Adrian Keister, clathratus, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    is the hypothesis true for every $theta$?
    $endgroup$
    – mookid
    Sep 19 '14 at 15:33














  • 1




    $begingroup$
    is the hypothesis true for every $theta$?
    $endgroup$
    – mookid
    Sep 19 '14 at 15:33








1




1




$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33




$begingroup$
is the hypothesis true for every $theta$?
$endgroup$
– mookid
Sep 19 '14 at 15:33










2 Answers
2






active

oldest

votes


















5












$begingroup$

This problem can be found in




LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.




The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.




I will write here the solution based on the second method:



begin{align*}
(sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
&= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
& leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
&= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
&= (mcos^2{theta} + nsin^2{theta})\
&<l.
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
    $endgroup$
    – user62029
    Sep 19 '14 at 16:16












  • $begingroup$
    You are welcome.
    $endgroup$
    – user161303
    Sep 19 '14 at 16:22



















0












$begingroup$

Hint: write
$$
sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
$$
and look into the Cauchy-Schwarz inequality.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    This problem can be found in




    LARSON, LOREN C., 1983: Problem-Solving Through Problems.
    Springer-Verlag, p. 255, 267.




    The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.




    I will write here the solution based on the second method:



    begin{align*}
    (sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
    &= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    & leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    &= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
    &= (mcos^2{theta} + nsin^2{theta})\
    &<l.
    end{align*}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
      $endgroup$
      – user62029
      Sep 19 '14 at 16:16












    • $begingroup$
      You are welcome.
      $endgroup$
      – user161303
      Sep 19 '14 at 16:22
















    5












    $begingroup$

    This problem can be found in




    LARSON, LOREN C., 1983: Problem-Solving Through Problems.
    Springer-Verlag, p. 255, 267.




    The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.




    I will write here the solution based on the second method:



    begin{align*}
    (sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
    &= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    & leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    &= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
    &= (mcos^2{theta} + nsin^2{theta})\
    &<l.
    end{align*}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
      $endgroup$
      – user62029
      Sep 19 '14 at 16:16












    • $begingroup$
      You are welcome.
      $endgroup$
      – user161303
      Sep 19 '14 at 16:22














    5












    5








    5





    $begingroup$

    This problem can be found in




    LARSON, LOREN C., 1983: Problem-Solving Through Problems.
    Springer-Verlag, p. 255, 267.




    The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.




    I will write here the solution based on the second method:



    begin{align*}
    (sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
    &= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    & leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    &= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
    &= (mcos^2{theta} + nsin^2{theta})\
    &<l.
    end{align*}






    share|cite|improve this answer









    $endgroup$



    This problem can be found in




    LARSON, LOREN C., 1983: Problem-Solving Through Problems.
    Springer-Verlag, p. 255, 267.




    The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavity of $ f(x) = sqrt{x} $, is left as an exercise.




    I will write here the solution based on the second method:



    begin{align*}
    (sqrt{m}cos^2{theta} + sqrt{n}sin^2{theta})^2
    &= m cos^4{theta} + 2 sqrt{mn}cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    & leq m cos^4{theta} + (m + n)cos^2{theta}sin^2{theta}+ nsin^4{theta} \
    &= (mcos^2{theta} + nsin^2{theta})(cos^2{theta} + sin^2{theta}) \
    &= (mcos^2{theta} + nsin^2{theta})\
    &<l.
    end{align*}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 19 '14 at 15:52







    user161303



















    • $begingroup$
      Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
      $endgroup$
      – user62029
      Sep 19 '14 at 16:16












    • $begingroup$
      You are welcome.
      $endgroup$
      – user161303
      Sep 19 '14 at 16:22


















    • $begingroup$
      Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
      $endgroup$
      – user62029
      Sep 19 '14 at 16:16












    • $begingroup$
      You are welcome.
      $endgroup$
      – user161303
      Sep 19 '14 at 16:22
















    $begingroup$
    Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
    $endgroup$
    – user62029
    Sep 19 '14 at 16:16






    $begingroup$
    Thank you for the reference and the hints. I'll try to solve the problem using all the methods.
    $endgroup$
    – user62029
    Sep 19 '14 at 16:16














    $begingroup$
    You are welcome.
    $endgroup$
    – user161303
    Sep 19 '14 at 16:22




    $begingroup$
    You are welcome.
    $endgroup$
    – user161303
    Sep 19 '14 at 16:22











    0












    $begingroup$

    Hint: write
    $$
    sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
    $$
    and look into the Cauchy-Schwarz inequality.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: write
      $$
      sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
      $$
      and look into the Cauchy-Schwarz inequality.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: write
        $$
        sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
        $$
        and look into the Cauchy-Schwarz inequality.






        share|cite|improve this answer









        $endgroup$



        Hint: write
        $$
        sqrt{m}cos^2theta+sqrt{n}sin^2theta=(sqrt{m}costheta)cdotcostheta+(sqrt{n}sintheta)cdotsintheta
        $$
        and look into the Cauchy-Schwarz inequality.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 19 '14 at 15:38









        Kim Jong UnKim Jong Un

        13.2k11738




        13.2k11738















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