Inequality with $(x+y)(y+z)(z+w)(w+x)=1$












4












$begingroup$


Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



I'm trying to use Holder's inequality
$$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
le (x+y)(y+z)(z+w)$$
$$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
so
$$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






inequality radicals a.m.-g.m.-inequality holder-inequality







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $x,y,z,w>0$ and such that
    $$(x+y)(y+z)(z+w)(w+x)=1.$$
    Show that
    $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



    I'm trying to use Holder's inequality
    $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
    le (x+y)(y+z)(z+w)$$
    $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
    so
    $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






    inequality radicals a.m.-g.m.-inequality holder-inequality







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      2



      $begingroup$


      Let $x,y,z,w>0$ and such that
      $$(x+y)(y+z)(z+w)(w+x)=1.$$
      Show that
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



      I'm trying to use Holder's inequality
      $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
      le (x+y)(y+z)(z+w)$$
      $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
      so
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






      inequality radicals a.m.-g.m.-inequality holder-inequality







      share|cite|improve this question











      $endgroup$




      Let $x,y,z,w>0$ and such that
      $$(x+y)(y+z)(z+w)(w+x)=1.$$
      Show that
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le 2.$$



      I'm trying to use Holder's inequality
      $$(sqrt[3]{xyz}+sqrt[3]{yzw})^3
      le (x+y)(y+z)(z+w)$$
      $$(sqrt[3]{zwx}+sqrt[3]{wxy})^3le (z+w)(w+x)(x+y)$$
      so
      $$sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}le sqrt[3]{(x+y)(y+z)(z+w)}+sqrt[3]{(z+w)(w+x)(x+y)}.$$






      inequality radicals a.m.-g.m.-inequality holder-inequality




      inequality radicals a.m.-g.m.-inequality holder-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 5:23









      Michael Rozenberg

      108k1895200




      108k1895200










      asked Dec 29 '18 at 1:19









      inequalityinequality

      785522




      785522






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you clarify the Holder part?
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            $begingroup$
            @Pratyush Sarkar I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            $begingroup$
            Ah, thanks that helps a lot. Nice concise answer.
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 6:56





















          1












          $begingroup$

          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice solution! +1
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 9:53













          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%2f3055447%2finequality-with-xyyzzwwx-1%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









          4












          $begingroup$

          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you clarify the Holder part?
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            $begingroup$
            @Pratyush Sarkar I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            $begingroup$
            Ah, thanks that helps a lot. Nice concise answer.
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 6:56


















          4












          $begingroup$

          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you clarify the Holder part?
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            $begingroup$
            @Pratyush Sarkar I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            $begingroup$
            Ah, thanks that helps a lot. Nice concise answer.
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 6:56
















          4












          4








          4





          $begingroup$

          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$






          share|cite|improve this answer











          $endgroup$



          Since $$prod_{cyc}(x+y)-sum_{cyc}xsum_{cyc}xyz=(xz-yw)^2geq0,$$ by Holder we obtain:



          $$2=2left(prod_{cyc}(x+y)right)^{frac{1}{4}}geq2left(sum_{cyc}xsum_{cyc}xyzright)^{frac{1}{4}}=$$
          $$=sqrt[4]{left(4^2sum_{cyc}xyzright)sum_{cyc}x}geqsqrt[4]{left(sum_{cyc}sqrt[3]{xyz}right)^3sum_{cyc}x}.$$
          Thus, it's enough to prove that
          $$sum_{cyc}xgeqsum_{cyc}sqrt[3]{xyz},$$ which is true by AM-GM:
          $$sum_{cyc}x=frac{1}{3}sum_{cyc}(x+y+z)geqfrac{1}{3}sum_{cyc}3sqrt[3]{xyz}=sum_{cyc}sqrt[3]{xyz}.$$
          Done!



          I used Holder for two sequences:
          $$16sum_{cyc}xyz=left(sum_{cyc}1right)^2sum_{cyc}xyzgeqleft(sum_{cyc}sqrt[3]{1^2xyz}right)^3=left(sum_{cyc}sqrt[3]{xyz}right)^3$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 6:45

























          answered Dec 29 '18 at 5:00









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200












          • $begingroup$
            Can you clarify the Holder part?
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            $begingroup$
            @Pratyush Sarkar I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            $begingroup$
            Ah, thanks that helps a lot. Nice concise answer.
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 6:56




















          • $begingroup$
            Can you clarify the Holder part?
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 5:51






          • 1




            $begingroup$
            @Pratyush Sarkar I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 29 '18 at 6:46






          • 1




            $begingroup$
            Ah, thanks that helps a lot. Nice concise answer.
            $endgroup$
            – Pratyush Sarkar
            Dec 29 '18 at 6:56


















          $begingroup$
          Can you clarify the Holder part?
          $endgroup$
          – Pratyush Sarkar
          Dec 29 '18 at 5:51




          $begingroup$
          Can you clarify the Holder part?
          $endgroup$
          – Pratyush Sarkar
          Dec 29 '18 at 5:51




          1




          1




          $begingroup$
          @Pratyush Sarkar I added something. See now.
          $endgroup$
          – Michael Rozenberg
          Dec 29 '18 at 6:46




          $begingroup$
          @Pratyush Sarkar I added something. See now.
          $endgroup$
          – Michael Rozenberg
          Dec 29 '18 at 6:46




          1




          1




          $begingroup$
          Ah, thanks that helps a lot. Nice concise answer.
          $endgroup$
          – Pratyush Sarkar
          Dec 29 '18 at 6:56






          $begingroup$
          Ah, thanks that helps a lot. Nice concise answer.
          $endgroup$
          – Pratyush Sarkar
          Dec 29 '18 at 6:56













          1












          $begingroup$

          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice solution! +1
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 9:53


















          1












          $begingroup$

          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very nice solution! +1
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 9:53
















          1












          1








          1





          $begingroup$

          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$






          share|cite|improve this answer









          $endgroup$



          This does not seem the best, but using only Holder's inequality, we can get
          $$begin{eqnarray}
          S^{12}&le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}left(prod_{text{cyc}}(x+y)right)^3 =2^{12},
          end{eqnarray}$$          for $S= sqrt[3]{xyz}+sqrt[3]{yzw}+sqrt[3]{zwx}+sqrt[3]{wxy}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 8:59









          SongSong

          18.3k21549




          18.3k21549












          • $begingroup$
            Very nice solution! +1
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 9:53




















          • $begingroup$
            Very nice solution! +1
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 9:53


















          $begingroup$
          Very nice solution! +1
          $endgroup$
          – Michael Rozenberg
          Dec 30 '18 at 9:53






          $begingroup$
          Very nice solution! +1
          $endgroup$
          – Michael Rozenberg
          Dec 30 '18 at 9:53




















          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%2f3055447%2finequality-with-xyyzzwwx-1%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

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more