A triangle has sides $a$, $b$, $c$ and medians $m_a$, $m_b$, $m_c$. Show...












5












$begingroup$



Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
$$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$




My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
$$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
I square it, but that doesn't help.






geometry inequality triangle geometric-inequalities sos







share|cite|improve this question











$endgroup$

















    5












    $begingroup$



    Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
    $$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$




    My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
    Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
    $$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
    right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
    I square it, but that doesn't help.






    geometry inequality triangle geometric-inequalities sos







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$



      Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
      $$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$




      My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
      Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
      $$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
      right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
      I square it, but that doesn't help.






      geometry inequality triangle geometric-inequalities sos







      share|cite|improve this question











      $endgroup$





      Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
      $$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$




      My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
      Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
      $$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
      right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
      I square it, but that doesn't help.






      geometry inequality triangle geometric-inequalities sos




      geometry inequality triangle geometric-inequalities sos






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 1:59









      Michael Rozenberg

      105k1892197




      105k1892197










      asked Dec 16 '18 at 12:14









      Winter In My Heart Winter In My Heart

      463




      463






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The hint:



          Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.



          Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
          $$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
          $$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
          Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
          $$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
          It est, it's enough to prove that
          $$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
          Now, let $a=y+z$, $b=x+z$ and $c=x+y$.



          Thus, $x$, $y$ and $z$ are positives and we need to prove that
          $$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
          $$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
          $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
          $$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
          Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.



          Thus,
          $$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
          $$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
          $$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
          $$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
          Done!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please help me prove the last. Do you have another solution?
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:29










          • $begingroup$
            @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
            $endgroup$
            – Michael Rozenberg
            Dec 16 '18 at 12:47










          • $begingroup$
            No problem, show it please
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:51










          • $begingroup$
            @Winter In My Heart I fixed a typo in my proof. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 17 '18 at 6:45











          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%2f3042546%2fa-triangle-has-sides-a-b-c-and-medians-m-a-m-b-m-c-show-abbc%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









          2












          $begingroup$

          The hint:



          Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.



          Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
          $$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
          $$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
          Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
          $$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
          It est, it's enough to prove that
          $$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
          Now, let $a=y+z$, $b=x+z$ and $c=x+y$.



          Thus, $x$, $y$ and $z$ are positives and we need to prove that
          $$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
          $$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
          $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
          $$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
          Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.



          Thus,
          $$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
          $$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
          $$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
          $$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
          Done!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please help me prove the last. Do you have another solution?
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:29










          • $begingroup$
            @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
            $endgroup$
            – Michael Rozenberg
            Dec 16 '18 at 12:47










          • $begingroup$
            No problem, show it please
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:51










          • $begingroup$
            @Winter In My Heart I fixed a typo in my proof. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 17 '18 at 6:45
















          2












          $begingroup$

          The hint:



          Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.



          Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
          $$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
          $$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
          Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
          $$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
          It est, it's enough to prove that
          $$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
          Now, let $a=y+z$, $b=x+z$ and $c=x+y$.



          Thus, $x$, $y$ and $z$ are positives and we need to prove that
          $$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
          $$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
          $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
          $$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
          Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.



          Thus,
          $$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
          $$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
          $$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
          $$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
          Done!






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Please help me prove the last. Do you have another solution?
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:29










          • $begingroup$
            @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
            $endgroup$
            – Michael Rozenberg
            Dec 16 '18 at 12:47










          • $begingroup$
            No problem, show it please
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:51










          • $begingroup$
            @Winter In My Heart I fixed a typo in my proof. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 17 '18 at 6:45














          2












          2








          2





          $begingroup$

          The hint:



          Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.



          Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
          $$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
          $$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
          Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
          $$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
          It est, it's enough to prove that
          $$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
          Now, let $a=y+z$, $b=x+z$ and $c=x+y$.



          Thus, $x$, $y$ and $z$ are positives and we need to prove that
          $$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
          $$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
          $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
          $$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
          Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.



          Thus,
          $$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
          $$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
          $$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
          $$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
          Done!






          share|cite|improve this answer











          $endgroup$



          The hint:



          Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.



          Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
          $$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
          $$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
          Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
          $$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
          It est, it's enough to prove that
          $$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
          Now, let $a=y+z$, $b=x+z$ and $c=x+y$.



          Thus, $x$, $y$ and $z$ are positives and we need to prove that
          $$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
          $$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
          $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
          $$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
          Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
          $$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.



          Thus,
          $$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
          $$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
          $$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
          $$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
          $$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
          Done!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 6:44

























          answered Dec 16 '18 at 12:26









          Michael RozenbergMichael Rozenberg

          105k1892197




          105k1892197












          • $begingroup$
            Please help me prove the last. Do you have another solution?
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:29










          • $begingroup$
            @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
            $endgroup$
            – Michael Rozenberg
            Dec 16 '18 at 12:47










          • $begingroup$
            No problem, show it please
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:51










          • $begingroup$
            @Winter In My Heart I fixed a typo in my proof. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 17 '18 at 6:45


















          • $begingroup$
            Please help me prove the last. Do you have another solution?
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:29










          • $begingroup$
            @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
            $endgroup$
            – Michael Rozenberg
            Dec 16 '18 at 12:47










          • $begingroup$
            No problem, show it please
            $endgroup$
            – Winter In My Heart
            Dec 16 '18 at 12:51










          • $begingroup$
            @Winter In My Heart I fixed a typo in my proof. See now.
            $endgroup$
            – Michael Rozenberg
            Dec 17 '18 at 6:45
















          $begingroup$
          Please help me prove the last. Do you have another solution?
          $endgroup$
          – Winter In My Heart
          Dec 16 '18 at 12:29




          $begingroup$
          Please help me prove the last. Do you have another solution?
          $endgroup$
          – Winter In My Heart
          Dec 16 '18 at 12:29












          $begingroup$
          @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
          $endgroup$
          – Michael Rozenberg
          Dec 16 '18 at 12:47




          $begingroup$
          @Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
          $endgroup$
          – Michael Rozenberg
          Dec 16 '18 at 12:47












          $begingroup$
          No problem, show it please
          $endgroup$
          – Winter In My Heart
          Dec 16 '18 at 12:51




          $begingroup$
          No problem, show it please
          $endgroup$
          – Winter In My Heart
          Dec 16 '18 at 12:51












          $begingroup$
          @Winter In My Heart I fixed a typo in my proof. See now.
          $endgroup$
          – Michael Rozenberg
          Dec 17 '18 at 6:45




          $begingroup$
          @Winter In My Heart I fixed a typo in my proof. See now.
          $endgroup$
          – Michael Rozenberg
          Dec 17 '18 at 6:45


















          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%2f3042546%2fa-triangle-has-sides-a-b-c-and-medians-m-a-m-b-m-c-show-abbc%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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more