$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$












0












$begingroup$


How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$



We also have $x^2+y^2+z^2 le 12$, if it's relevant.

I tried using AM-GM and AM-HM, but I can't get to that on the right side.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you really mean for the three terms on the left to be identical?
    $endgroup$
    – Steve Kass
    Dec 11 '18 at 14:22










  • $begingroup$
    It' a cyclic sum probably.
    $endgroup$
    – Oldboy
    Dec 11 '18 at 14:23










  • $begingroup$
    Oh. I edited it now. Sorry.
    $endgroup$
    – Pero
    Dec 11 '18 at 14:24






  • 1




    $begingroup$
    Engel's form of Cauchy Schwarz = Titu's lemma
    $endgroup$
    – achille hui
    Dec 11 '18 at 14:29










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    Dec 11 '18 at 14:34
















0












$begingroup$


How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$



We also have $x^2+y^2+z^2 le 12$, if it's relevant.

I tried using AM-GM and AM-HM, but I can't get to that on the right side.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you really mean for the three terms on the left to be identical?
    $endgroup$
    – Steve Kass
    Dec 11 '18 at 14:22










  • $begingroup$
    It' a cyclic sum probably.
    $endgroup$
    – Oldboy
    Dec 11 '18 at 14:23










  • $begingroup$
    Oh. I edited it now. Sorry.
    $endgroup$
    – Pero
    Dec 11 '18 at 14:24






  • 1




    $begingroup$
    Engel's form of Cauchy Schwarz = Titu's lemma
    $endgroup$
    – achille hui
    Dec 11 '18 at 14:29










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    Dec 11 '18 at 14:34














0












0








0


1



$begingroup$


How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$



We also have $x^2+y^2+z^2 le 12$, if it's relevant.

I tried using AM-GM and AM-HM, but I can't get to that on the right side.






algebra-precalculus







share|cite|improve this question











$endgroup$




How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$



We also have $x^2+y^2+z^2 le 12$, if it's relevant.

I tried using AM-GM and AM-HM, but I can't get to that on the right side.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 14:23







Pero

















asked Dec 11 '18 at 14:18









PeroPero

1327




1327












  • $begingroup$
    Do you really mean for the three terms on the left to be identical?
    $endgroup$
    – Steve Kass
    Dec 11 '18 at 14:22










  • $begingroup$
    It' a cyclic sum probably.
    $endgroup$
    – Oldboy
    Dec 11 '18 at 14:23










  • $begingroup$
    Oh. I edited it now. Sorry.
    $endgroup$
    – Pero
    Dec 11 '18 at 14:24






  • 1




    $begingroup$
    Engel's form of Cauchy Schwarz = Titu's lemma
    $endgroup$
    – achille hui
    Dec 11 '18 at 14:29










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    Dec 11 '18 at 14:34


















  • $begingroup$
    Do you really mean for the three terms on the left to be identical?
    $endgroup$
    – Steve Kass
    Dec 11 '18 at 14:22










  • $begingroup$
    It' a cyclic sum probably.
    $endgroup$
    – Oldboy
    Dec 11 '18 at 14:23










  • $begingroup$
    Oh. I edited it now. Sorry.
    $endgroup$
    – Pero
    Dec 11 '18 at 14:24






  • 1




    $begingroup$
    Engel's form of Cauchy Schwarz = Titu's lemma
    $endgroup$
    – achille hui
    Dec 11 '18 at 14:29










  • $begingroup$
    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.
    $endgroup$
    – Carl Mummert
    Dec 11 '18 at 14:34
















$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22




$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22












$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23




$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23












$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24




$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24




1




1




$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29




$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29












$begingroup$
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.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34




$begingroup$
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.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34










2 Answers
2






active

oldest

votes


















3












$begingroup$

$AMge HM$



$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$



$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$



Multiply both sides by $6$ and you have your solution.



$[(sqrt6+sqrt6+sqrt6)^2=54]$






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
    $$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
    3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
    Now
    $$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
    frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
    frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
    which is exactly your RHS.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      ...which is Jensen.
      $endgroup$
      – Oldboy
      Dec 11 '18 at 14:26










    • $begingroup$
      Yes, by Jensen's inequality.
      $endgroup$
      – Robert Z
      Dec 11 '18 at 14:27











    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%2f3035329%2ffrac610x2y2-frac610y2z2-frac610x2z2-ge-frac-sqrt6-sqrt%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












    $begingroup$

    $AMge HM$



    $displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$



    $displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$



    Multiply both sides by $6$ and you have your solution.



    $[(sqrt6+sqrt6+sqrt6)^2=54]$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $AMge HM$



      $displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$



      $displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$



      Multiply both sides by $6$ and you have your solution.



      $[(sqrt6+sqrt6+sqrt6)^2=54]$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $AMge HM$



        $displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$



        $displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$



        Multiply both sides by $6$ and you have your solution.



        $[(sqrt6+sqrt6+sqrt6)^2=54]$






        share|cite|improve this answer











        $endgroup$



        $AMge HM$



        $displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$



        $displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$



        Multiply both sides by $6$ and you have your solution.



        $[(sqrt6+sqrt6+sqrt6)^2=54]$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 14:54

























        answered Dec 11 '18 at 14:30









        Shubham JohriShubham Johri

        5,082717




        5,082717























            3












            $begingroup$

            Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
            $$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
            3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
            Now
            $$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
            frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
            frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
            which is exactly your RHS.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ...which is Jensen.
              $endgroup$
              – Oldboy
              Dec 11 '18 at 14:26










            • $begingroup$
              Yes, by Jensen's inequality.
              $endgroup$
              – Robert Z
              Dec 11 '18 at 14:27
















            3












            $begingroup$

            Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
            $$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
            3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
            Now
            $$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
            frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
            frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
            which is exactly your RHS.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ...which is Jensen.
              $endgroup$
              – Oldboy
              Dec 11 '18 at 14:26










            • $begingroup$
              Yes, by Jensen's inequality.
              $endgroup$
              – Robert Z
              Dec 11 '18 at 14:27














            3












            3








            3





            $begingroup$

            Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
            $$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
            3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
            Now
            $$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
            frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
            frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
            which is exactly your RHS.






            share|cite|improve this answer











            $endgroup$



            Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
            $$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
            3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
            Now
            $$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
            frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
            frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
            which is exactly your RHS.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 11 '18 at 14:32

























            answered Dec 11 '18 at 14:25









            Robert ZRobert Z

            96.1k1065136




            96.1k1065136












            • $begingroup$
              ...which is Jensen.
              $endgroup$
              – Oldboy
              Dec 11 '18 at 14:26










            • $begingroup$
              Yes, by Jensen's inequality.
              $endgroup$
              – Robert Z
              Dec 11 '18 at 14:27


















            • $begingroup$
              ...which is Jensen.
              $endgroup$
              – Oldboy
              Dec 11 '18 at 14:26










            • $begingroup$
              Yes, by Jensen's inequality.
              $endgroup$
              – Robert Z
              Dec 11 '18 at 14:27
















            $begingroup$
            ...which is Jensen.
            $endgroup$
            – Oldboy
            Dec 11 '18 at 14:26




            $begingroup$
            ...which is Jensen.
            $endgroup$
            – Oldboy
            Dec 11 '18 at 14:26












            $begingroup$
            Yes, by Jensen's inequality.
            $endgroup$
            – Robert Z
            Dec 11 '18 at 14:27




            $begingroup$
            Yes, by Jensen's inequality.
            $endgroup$
            – Robert Z
            Dec 11 '18 at 14:27


















            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%2f3035329%2ffrac610x2y2-frac610y2z2-frac610x2z2-ge-frac-sqrt6-sqrt%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

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more