Minimum $M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$ for $3x^2+8y^3=20$












0












$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55
















0












$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55














0












0








0


1



$begingroup$



Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima







share|cite|improve this question











$endgroup$





Let $x,y$ be positive real numbers such that $xneq y$ and
$3x^{2}+8y^{3}=20$. Find $x,y$ so that $M$ get the minimum value:
$M=frac{4}{x^{2}}+frac{4}{y^{2}}+frac{1}{(x-y)^{2}}$.




I can not find when this function gets the minimum value. Or can this difficult problem be solved by differentiating? How do we turn the root to another specie which has $ax+by$ or $frac{ax}{y}$ while $x,y$ are some constants.



Help me to solve this problem.






calculus inequality optimization nonlinear-optimization maxima-minima




calculus inequality optimization nonlinear-optimization maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:27









Martin Sleziak

44.8k10119272




44.8k10119272










asked Dec 13 '18 at 11:26









Trong TuanTrong Tuan

1318




1318








  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55














  • 1




    $begingroup$
    What does > mean in the second line?
    $endgroup$
    – William Elliot
    Dec 13 '18 at 11:55








1




1




$begingroup$
What does > mean in the second line?
$endgroup$
– William Elliot
Dec 13 '18 at 11:55




$begingroup$
What does > mean in the second line?
$endgroup$
– William Elliot
Dec 13 '18 at 11:55










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $x=2$ and $y=1$.



Thus, we'll get a value $6$.



We'll prove that it's a minimal value.



Indeed, let $x=2a$ and $y=b$.



Thus, $3a^2+2b^3=5$ and we need to prove that
$$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
But by AM-GM
$$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
$$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
which gives $2geq a^2+b^2.$



Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
Now, let $a=tb$.



Thus, we need to prove that
$$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
$$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



But for $0<tleq1$ we obtain:
$$16t^4+16t^3-7t^2-2t+1=$$
$$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
Done!



Also, we can use derivative here.



Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
$$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
$$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28










  • $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50










  • $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38





















0












$begingroup$

Local minimum:
$$f(2,1)=6$$
Global minimum:
$$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
enter image description here






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037891%2fminimum-m-frac4x2-frac4y2-frac1x-y2-for-3x28y3-2%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38


















    3












    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38
















    3












    3








    3





    $begingroup$

    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.






    share|cite|improve this answer











    $endgroup$



    Let $x=2$ and $y=1$.



    Thus, we'll get a value $6$.



    We'll prove that it's a minimal value.



    Indeed, let $x=2a$ and $y=b$.



    Thus, $3a^2+2b^3=5$ and we need to prove that
    $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geq6.$$
    But by AM-GM
    $$5=3a^2+2b^3=3a^2-1+3b^2+2b^3+1-3b^2geq$$
    $$geq3a^2+3b^2-1+3sqrt[3]{(b^3)^2cdot1}-3b^2=3(a^2+b^2)-1,$$
    which gives $2geq a^2+b^2.$



    Id est, $$frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}geqfrac{1}{2}(a^2+b^2)left(frac{1}{a^2}+frac{4}{b^2}+frac{1}{(2a-b)^2}right).$$
    Now, let $a=tb$.



    Thus, we need to prove that
    $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ or
    $$(t-1)^2(16t^4+16t^3-7t^2-2t+1)geq0,$$ which is obviously true for $tgeq1.$



    But for $0<tleq1$ we obtain:
    $$16t^4+16t^3-7t^2-2t+1=$$
    $$=9t^3(1-t)+left(5t^2+frac{7}{10}t-frac{5}{6}right)^2+frac{1}{900}(759t^2-750t+275)geq0.$$
    Done!



    Also, we can use derivative here.



    Since $$(t^2+1)left(frac{1}{t^2}+4+frac{1}{(2t-1)^2}right)geq12$$ it's $f(t)geq0,$ where
    $$f(t)=64t^6-64t^5-92t^4+112t^3-8t^2-16t+4,$$ and
    $$f'(t)=(t-1)(t+1)(3t-1)(8t^2-4t-1),$$
    we obtain $t_{min}=frac{1}{3}$ or $t_{min}=1$ and since $fleft(frac{1}{3}right)>f(1)=0$, we are done.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 13:49

























    answered Dec 13 '18 at 12:04









    Michael RozenbergMichael Rozenberg

    106k1893198




    106k1893198












    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38




















    • $begingroup$
      Can w use derivative to solve this. Thank you
      $endgroup$
      – Trong Tuan
      Dec 13 '18 at 13:28










    • $begingroup$
      @Trong Tuan I added something. See now.
      $endgroup$
      – Michael Rozenberg
      Dec 13 '18 at 13:50










    • $begingroup$
      In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
      $endgroup$
      – Aleksas Domarkas
      Dec 19 '18 at 11:38


















    $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28




    $begingroup$
    Can w use derivative to solve this. Thank you
    $endgroup$
    – Trong Tuan
    Dec 13 '18 at 13:28












    $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50




    $begingroup$
    @Trong Tuan I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Dec 13 '18 at 13:50












    $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38






    $begingroup$
    In $$x=2,; y=1$$ is local minimum. Global minimum is in $x=-1.668453756732446,;y=1.133435720989714$
    $endgroup$
    – Aleksas Domarkas
    Dec 19 '18 at 11:38













    0












    $begingroup$

    Local minimum:
    $$f(2,1)=6$$
    Global minimum:
    $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Local minimum:
      $$f(2,1)=6$$
      Global minimum:
      $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Local minimum:
        $$f(2,1)=6$$
        Global minimum:
        $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
        enter image description here






        share|cite|improve this answer









        $endgroup$



        Local minimum:
        $$f(2,1)=6$$
        Global minimum:
        $$f(−1.668453756732446,1.133435720989714)=4.677920159716602$$
        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 20:43









        Aleksas DomarkasAleksas Domarkas

        1,37216




        1,37216






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037891%2fminimum-m-frac4x2-frac4y2-frac1x-y2-for-3x28y3-2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more