geometry inequality












-1












$begingroup$


$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.



Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$






geometry inequality triangle geometric-inequalities







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  • 1




    $begingroup$
    Hint: consider areas and use AM-GM.
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:49












  • $begingroup$
    What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
    $endgroup$
    – jayant98
    Dec 9 '18 at 9:52
















-1












$begingroup$


$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.



Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$






geometry inequality triangle geometric-inequalities







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: consider areas and use AM-GM.
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:49












  • $begingroup$
    What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
    $endgroup$
    – jayant98
    Dec 9 '18 at 9:52














-1












-1








-1


1



$begingroup$


$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.



Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$






geometry inequality triangle geometric-inequalities







share|cite|improve this question











$endgroup$




$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.



Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$






geometry inequality triangle geometric-inequalities




geometry inequality triangle geometric-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 9 '18 at 12:54









Martin Sleziak

44.7k9117272




44.7k9117272










asked Dec 9 '18 at 9:42









Nguyen Xuan MinhNguyen Xuan Minh

62




62








  • 1




    $begingroup$
    Hint: consider areas and use AM-GM.
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:49












  • $begingroup$
    What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
    $endgroup$
    – jayant98
    Dec 9 '18 at 9:52














  • 1




    $begingroup$
    Hint: consider areas and use AM-GM.
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:49












  • $begingroup$
    What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
    $endgroup$
    – jayant98
    Dec 9 '18 at 9:52








1




1




$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49






$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49














$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52




$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52










3 Answers
3






active

oldest

votes


















0












$begingroup$

$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:



    $27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
      $endgroup$
      – José Carlos Santos
      Dec 9 '18 at 10:07



















    0












    $begingroup$

    By AM-GM
    $$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
    $$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$






          share|cite|improve this answer









          $endgroup$



          $$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 10:07









          Sergic PrimazonSergic Primazon

          54215




          54215























              0












              $begingroup$

              Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:



              $27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                $endgroup$
                – José Carlos Santos
                Dec 9 '18 at 10:07
















              0












              $begingroup$

              Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:



              $27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                $endgroup$
                – José Carlos Santos
                Dec 9 '18 at 10:07














              0












              0








              0





              $begingroup$

              Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:



              $27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.






              share|cite|improve this answer











              $endgroup$



              Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:



              $27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 9 '18 at 10:36









              Anubhab Ghosal

              82518




              82518










              answered Dec 9 '18 at 9:51









              Ziqing XUZiqing XU

              11




              11












              • $begingroup$
                Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                $endgroup$
                – José Carlos Santos
                Dec 9 '18 at 10:07


















              • $begingroup$
                Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                $endgroup$
                – José Carlos Santos
                Dec 9 '18 at 10:07
















              $begingroup$
              Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 10:07




              $begingroup$
              Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
              $endgroup$
              – José Carlos Santos
              Dec 9 '18 at 10:07











              0












              $begingroup$

              By AM-GM
              $$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
              $$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By AM-GM
                $$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
                $$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By AM-GM
                  $$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
                  $$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$






                  share|cite|improve this answer









                  $endgroup$



                  By AM-GM
                  $$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
                  $$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 10:42









                  Michael RozenbergMichael Rozenberg

                  98.7k1590189




                  98.7k1590189






























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