Prove that if $a,b,c>0$ then ${1over a}+{1over b}+{1over c}ge{2over a+b}+{2over b+c}+{2over c+a}$...












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This question already has an answer here:




  • An inequality concerning three positive numbers

    1 answer





Prove that if $a,b,c>0$ then ${1over a}+{1over b}+{1over c}ge{2over a+b}+{2over b+c}+{2over c+a}$




Excuse me but I am new to inequalities and have been practicing some questions including the one above. The one above looks easy but for me is not. I tried to use AM-GM-HM but led to even greater problems. Here is my attempt:
$${a+b+cover3}ge{3over{1over a}+{1over b}+{1over c}}$$
$${1over 2}biggl({(a+b)+(b+c)+(c+a)over3}biggr)ge{3over{1over a}+{1over b}+{1over c}}$$
$$iff {{1over a}+{1over b}+{1over c}over3}geBiggl({1over{6over a+b}+{6over b+c}+{6over c+a}}Biggr)$$
I could barely spot the wanted up there but cannot continue further.
I also got that
$${2over a+b}+{2over b+c}+{2over c+a}ge{1over a+b+c}$$
Any help, if maybe a different method, would be appreciated. Thank you in advance.






algebra-precalculus







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marked as duplicate by Martin R, clathratus, Jyrki Lahtonen, amWhy algebra-precalculus
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 Jan 1 at 21:12


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
    $endgroup$
    – Love Invariants
    Jan 1 at 19:02












  • $begingroup$
    @LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
    $endgroup$
    – user587054
    Jan 1 at 19:04








  • 1




    $begingroup$
    I'm talking about the question though. I think it is wrong.
    $endgroup$
    – Love Invariants
    Jan 1 at 19:07










  • $begingroup$
    Oh it's written wrong than you.
    $endgroup$
    – user587054
    Jan 1 at 19:08










  • $begingroup$
    It is fixed now
    $endgroup$
    – user587054
    Jan 1 at 19:09
















1












$begingroup$



This question already has an answer here:




  • An inequality concerning three positive numbers

    1 answer





Prove that if $a,b,c>0$ then ${1over a}+{1over b}+{1over c}ge{2over a+b}+{2over b+c}+{2over c+a}$




Excuse me but I am new to inequalities and have been practicing some questions including the one above. The one above looks easy but for me is not. I tried to use AM-GM-HM but led to even greater problems. Here is my attempt:
$${a+b+cover3}ge{3over{1over a}+{1over b}+{1over c}}$$
$${1over 2}biggl({(a+b)+(b+c)+(c+a)over3}biggr)ge{3over{1over a}+{1over b}+{1over c}}$$
$$iff {{1over a}+{1over b}+{1over c}over3}geBiggl({1over{6over a+b}+{6over b+c}+{6over c+a}}Biggr)$$
I could barely spot the wanted up there but cannot continue further.
I also got that
$${2over a+b}+{2over b+c}+{2over c+a}ge{1over a+b+c}$$
Any help, if maybe a different method, would be appreciated. Thank you in advance.






algebra-precalculus







share|cite|improve this question











$endgroup$



marked as duplicate by Martin R, clathratus, Jyrki Lahtonen, amWhy algebra-precalculus
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 Jan 1 at 21:12


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
    $endgroup$
    – Love Invariants
    Jan 1 at 19:02












  • $begingroup$
    @LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
    $endgroup$
    – user587054
    Jan 1 at 19:04








  • 1




    $begingroup$
    I'm talking about the question though. I think it is wrong.
    $endgroup$
    – Love Invariants
    Jan 1 at 19:07










  • $begingroup$
    Oh it's written wrong than you.
    $endgroup$
    – user587054
    Jan 1 at 19:08










  • $begingroup$
    It is fixed now
    $endgroup$
    – user587054
    Jan 1 at 19:09














1












1








1





$begingroup$



This question already has an answer here:




  • An inequality concerning three positive numbers

    1 answer





Prove that if $a,b,c>0$ then ${1over a}+{1over b}+{1over c}ge{2over a+b}+{2over b+c}+{2over c+a}$




Excuse me but I am new to inequalities and have been practicing some questions including the one above. The one above looks easy but for me is not. I tried to use AM-GM-HM but led to even greater problems. Here is my attempt:
$${a+b+cover3}ge{3over{1over a}+{1over b}+{1over c}}$$
$${1over 2}biggl({(a+b)+(b+c)+(c+a)over3}biggr)ge{3over{1over a}+{1over b}+{1over c}}$$
$$iff {{1over a}+{1over b}+{1over c}over3}geBiggl({1over{6over a+b}+{6over b+c}+{6over c+a}}Biggr)$$
I could barely spot the wanted up there but cannot continue further.
I also got that
$${2over a+b}+{2over b+c}+{2over c+a}ge{1over a+b+c}$$
Any help, if maybe a different method, would be appreciated. Thank you in advance.






algebra-precalculus







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • An inequality concerning three positive numbers

    1 answer





Prove that if $a,b,c>0$ then ${1over a}+{1over b}+{1over c}ge{2over a+b}+{2over b+c}+{2over c+a}$




Excuse me but I am new to inequalities and have been practicing some questions including the one above. The one above looks easy but for me is not. I tried to use AM-GM-HM but led to even greater problems. Here is my attempt:
$${a+b+cover3}ge{3over{1over a}+{1over b}+{1over c}}$$
$${1over 2}biggl({(a+b)+(b+c)+(c+a)over3}biggr)ge{3over{1over a}+{1over b}+{1over c}}$$
$$iff {{1over a}+{1over b}+{1over c}over3}geBiggl({1over{6over a+b}+{6over b+c}+{6over c+a}}Biggr)$$
I could barely spot the wanted up there but cannot continue further.
I also got that
$${2over a+b}+{2over b+c}+{2over c+a}ge{1over a+b+c}$$
Any help, if maybe a different method, would be appreciated. Thank you in advance.





This question already has an answer here:




  • An inequality concerning three positive numbers

    1 answer







algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 19:09







user587054

















asked Jan 1 at 18:58









user587054user587054

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46511




marked as duplicate by Martin R, clathratus, Jyrki Lahtonen, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

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 Jan 1 at 21:12


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Martin R, clathratus, Jyrki Lahtonen, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

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 Jan 1 at 21:12


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
    $endgroup$
    – Love Invariants
    Jan 1 at 19:02












  • $begingroup$
    @LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
    $endgroup$
    – user587054
    Jan 1 at 19:04








  • 1




    $begingroup$
    I'm talking about the question though. I think it is wrong.
    $endgroup$
    – Love Invariants
    Jan 1 at 19:07










  • $begingroup$
    Oh it's written wrong than you.
    $endgroup$
    – user587054
    Jan 1 at 19:08










  • $begingroup$
    It is fixed now
    $endgroup$
    – user587054
    Jan 1 at 19:09














  • 1




    $begingroup$
    What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
    $endgroup$
    – Love Invariants
    Jan 1 at 19:02












  • $begingroup$
    @LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
    $endgroup$
    – user587054
    Jan 1 at 19:04








  • 1




    $begingroup$
    I'm talking about the question though. I think it is wrong.
    $endgroup$
    – Love Invariants
    Jan 1 at 19:07










  • $begingroup$
    Oh it's written wrong than you.
    $endgroup$
    – user587054
    Jan 1 at 19:08










  • $begingroup$
    It is fixed now
    $endgroup$
    – user587054
    Jan 1 at 19:09








1




1




$begingroup$
What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
$endgroup$
– Love Invariants
Jan 1 at 19:02






$begingroup$
What if you plug a=b=c=2? The first and second both inequalities would be invalid. I think it is for $a,b,c le 1$
$endgroup$
– Love Invariants
Jan 1 at 19:02














$begingroup$
@LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
$endgroup$
– user587054
Jan 1 at 19:04






$begingroup$
@LoveInvariants yes I made a mistake, its meant to be 6 instead of 3.
$endgroup$
– user587054
Jan 1 at 19:04






1




1




$begingroup$
I'm talking about the question though. I think it is wrong.
$endgroup$
– Love Invariants
Jan 1 at 19:07




$begingroup$
I'm talking about the question though. I think it is wrong.
$endgroup$
– Love Invariants
Jan 1 at 19:07












$begingroup$
Oh it's written wrong than you.
$endgroup$
– user587054
Jan 1 at 19:08




$begingroup$
Oh it's written wrong than you.
$endgroup$
– user587054
Jan 1 at 19:08












$begingroup$
It is fixed now
$endgroup$
– user587054
Jan 1 at 19:09




$begingroup$
It is fixed now
$endgroup$
– user587054
Jan 1 at 19:09










1 Answer
1






active

oldest

votes


















2












$begingroup$

We apply the arithmetic-harmonic mean inequality in variables $a,b$:



$$frac{1}{2}left(frac{1}{a}+frac{1}{b}right)ge frac{2}{1/(1/a)+1/(1/b)}=frac{2}{a+b}$$



Now do this twice more: in variables $a,c$, and in variables $b,c$. Then add the three inequalities.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We apply the arithmetic-harmonic mean inequality in variables $a,b$:



    $$frac{1}{2}left(frac{1}{a}+frac{1}{b}right)ge frac{2}{1/(1/a)+1/(1/b)}=frac{2}{a+b}$$



    Now do this twice more: in variables $a,c$, and in variables $b,c$. Then add the three inequalities.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      We apply the arithmetic-harmonic mean inequality in variables $a,b$:



      $$frac{1}{2}left(frac{1}{a}+frac{1}{b}right)ge frac{2}{1/(1/a)+1/(1/b)}=frac{2}{a+b}$$



      Now do this twice more: in variables $a,c$, and in variables $b,c$. Then add the three inequalities.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        We apply the arithmetic-harmonic mean inequality in variables $a,b$:



        $$frac{1}{2}left(frac{1}{a}+frac{1}{b}right)ge frac{2}{1/(1/a)+1/(1/b)}=frac{2}{a+b}$$



        Now do this twice more: in variables $a,c$, and in variables $b,c$. Then add the three inequalities.






        share|cite|improve this answer









        $endgroup$



        We apply the arithmetic-harmonic mean inequality in variables $a,b$:



        $$frac{1}{2}left(frac{1}{a}+frac{1}{b}right)ge frac{2}{1/(1/a)+1/(1/b)}=frac{2}{a+b}$$



        Now do this twice more: in variables $a,c$, and in variables $b,c$. Then add the three inequalities.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 19:12









        vadim123vadim123

        75.7k897189




        75.7k897189















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