Prove logarithmic inequality:












1












$begingroup$


Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



(It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
$$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



I can get the left side to
$$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



But I don't know how to get the right side as in the solution. How do I get that result?






algebra-precalculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



    (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



    I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
    $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



    I can get the left side to
    $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



    But I don't know how to get the right side as in the solution. How do I get that result?






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



      (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



      I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
      $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



      I can get the left side to
      $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



      But I don't know how to get the right side as in the solution. How do I get that result?






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



      (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



      I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
      $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



      I can get the left side to
      $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



      But I don't know how to get the right side as in the solution. How do I get that result?






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 22:35









      PeroPero

      1277




      1277






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19











          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%2f3034588%2fprove-logarithmic-inequality%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









          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19
















          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19














          1












          1








          1





          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$



          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 22:38









          gimusigimusi

          92.8k84494




          92.8k84494












          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19


















          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19
















          $begingroup$
          Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
          $endgroup$
          – Pero
          Dec 10 '18 at 22:58






          $begingroup$
          Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
          $endgroup$
          – Pero
          Dec 10 '18 at 22:58














          $begingroup$
          Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:05




          $begingroup$
          Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:05












          $begingroup$
          We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:11






          $begingroup$
          We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:11














          $begingroup$
          @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:13






          $begingroup$
          @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:13














          $begingroup$
          Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:19




          $begingroup$
          Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:19


















          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%2f3034588%2fprove-logarithmic-inequality%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          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

          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