Prove $frac{a+b}{sqrt{c}}+frac{a+c}{sqrt{b}}+ frac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c})$












3














Could anyone advise me on how to prove this inequality:



$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$



where $a,b,c $ are any positive real numbers.



Do I use the AM-GM inequality somewhere?



Thank you.










share|cite|improve this question
























  • Sorry, I made a typo error. It should be $geq$.
    – Alexy Vincenzo
    Nov 4 '15 at 1:01
















3














Could anyone advise me on how to prove this inequality:



$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$



where $a,b,c $ are any positive real numbers.



Do I use the AM-GM inequality somewhere?



Thank you.










share|cite|improve this question
























  • Sorry, I made a typo error. It should be $geq$.
    – Alexy Vincenzo
    Nov 4 '15 at 1:01














3












3








3


1





Could anyone advise me on how to prove this inequality:



$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$



where $a,b,c $ are any positive real numbers.



Do I use the AM-GM inequality somewhere?



Thank you.










share|cite|improve this question















Could anyone advise me on how to prove this inequality:



$$dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} geq 2(sqrt{a} + sqrt{b} +sqrt{c}),$$



where $a,b,c $ are any positive real numbers.



Do I use the AM-GM inequality somewhere?



Thank you.







algebra-precalculus inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 16:44









Martin Sleziak

44.6k8115271




44.6k8115271










asked Nov 4 '15 at 0:54









Alexy VincenzoAlexy Vincenzo

2,1753925




2,1753925












  • Sorry, I made a typo error. It should be $geq$.
    – Alexy Vincenzo
    Nov 4 '15 at 1:01


















  • Sorry, I made a typo error. It should be $geq$.
    – Alexy Vincenzo
    Nov 4 '15 at 1:01
















Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01




Sorry, I made a typo error. It should be $geq$.
– Alexy Vincenzo
Nov 4 '15 at 1:01










1 Answer
1






active

oldest

votes


















7














From AM-GM you have



$$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
and
$$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
Doing this for the other variables and summing respectively you get
$$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
Simplifying gives the desired inequality.






share|cite|improve this answer























    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%2f1512074%2fprove-fracab-sqrtc-fracac-sqrtb-fracbc-sqrta-geq-2-s%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    From AM-GM you have



    $$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
    and
    $$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
    Doing this for the other variables and summing respectively you get
    $$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
    Simplifying gives the desired inequality.






    share|cite|improve this answer




























      7














      From AM-GM you have



      $$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
      and
      $$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
      Doing this for the other variables and summing respectively you get
      $$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
      Simplifying gives the desired inequality.






      share|cite|improve this answer


























        7












        7








        7






        From AM-GM you have



        $$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
        and
        $$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
        Doing this for the other variables and summing respectively you get
        $$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
        Simplifying gives the desired inequality.






        share|cite|improve this answer














        From AM-GM you have



        $$ frac{a}{sqrt{b}} +frac{a}{sqrt{b}} +frac{b}{sqrt{a}} geq 3 sqrt{a} $$
        and
        $$ frac{a}{sqrt{c}} +frac{a}{sqrt{c}} +frac{c}{sqrt{a}} geq 3 sqrt{a} $$
        Doing this for the other variables and summing respectively you get
        $$3left ( dfrac{a+b}{sqrt{c}}+dfrac{a+c}{sqrt{b}}+ dfrac{b+c}{sqrt{a}} right) geq 6( sqrt{a} + sqrt{b} +sqrt{c} )$$
        Simplifying gives the desired inequality.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 4 '15 at 1:28









        Kaster

        9,04221729




        9,04221729










        answered Nov 4 '15 at 1:25









        clarkclark

        12.9k22657




        12.9k22657






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f1512074%2fprove-fracab-sqrtc-fracac-sqrtb-fracbc-sqrta-geq-2-s%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

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅