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












5












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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











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    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

      98.5k1590189




      98.5k1590189










      asked Dec 16 '18 at 12:14









      Winter In My Heart Winter In My Heart

      463




      463






















          1 Answer
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          active

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          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











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          $begingroup$

          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

          98.5k1590189




          98.5k1590189












          • $begingroup$
            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


















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