Algebraic inequality for positive reals $a,b,c$












0












$begingroup$


The problem is from a previous maths olympiad and the last step is to prove the inequality




$$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 +4abc^4 geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $a,b,c > 0$.




I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you






inequality a.m.-g.m.-inequality rearrangement-inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
    $endgroup$
    – Song
    Jan 5 at 12:38






  • 2




    $begingroup$
    It would be better to see the original problem.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 12:45










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Jan 11 at 13:54
















0












$begingroup$


The problem is from a previous maths olympiad and the last step is to prove the inequality




$$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 +4abc^4 geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $a,b,c > 0$.




I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you






inequality a.m.-g.m.-inequality rearrangement-inequality







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
    $endgroup$
    – Song
    Jan 5 at 12:38






  • 2




    $begingroup$
    It would be better to see the original problem.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 12:45










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Jan 11 at 13:54














0












0








0





$begingroup$


The problem is from a previous maths olympiad and the last step is to prove the inequality




$$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 +4abc^4 geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $a,b,c > 0$.




I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you






inequality a.m.-g.m.-inequality rearrangement-inequality







share|cite|improve this question











$endgroup$




The problem is from a previous maths olympiad and the last step is to prove the inequality




$$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 +4abc^4 geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $a,b,c > 0$.




I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you






inequality a.m.-g.m.-inequality rearrangement-inequality




inequality a.m.-g.m.-inequality rearrangement-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 2:55









Michael Rozenberg

107k1894198




107k1894198










asked Jan 5 at 12:34









MathFanaticsMathFanatics

1769




1769








  • 2




    $begingroup$
    Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
    $endgroup$
    – Song
    Jan 5 at 12:38






  • 2




    $begingroup$
    It would be better to see the original problem.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 12:45










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Jan 11 at 13:54














  • 2




    $begingroup$
    Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
    $endgroup$
    – Song
    Jan 5 at 12:38






  • 2




    $begingroup$
    It would be better to see the original problem.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 12:45










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Jan 11 at 13:54








2




2




$begingroup$
Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
$endgroup$
– Song
Jan 5 at 12:38




$begingroup$
Isn't $a^4c$ a typo? I guess it should be $a^4c^2$.
$endgroup$
– Song
Jan 5 at 12:38




2




2




$begingroup$
It would be better to see the original problem.
$endgroup$
– Dr. Sonnhard Graubner
Jan 5 at 12:45




$begingroup$
It would be better to see the original problem.
$endgroup$
– Dr. Sonnhard Graubner
Jan 5 at 12:45












$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:54




$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:54










2 Answers
2






active

oldest

votes


















0












$begingroup$

The hint:



You can use AM-GM and Rearrangement.



For example, by Rearrangement
$$sum_{cyc}(a^4bc-a^3b^2c)=abcsum_{cyc}(a^3-a^2b)geq0.$$
Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.



Thus, $$sum_{cyc}a^3=a^2cdot a+b^2cdot b+c^2cdot cgeq a^2cdot b+b^2cdot c+c^2cdot a=sum_{cyc}a^2b.$$



Now, by AM-GM
$$sum_{cyc}(a^3b^3+a^3c^2b)geqsum_{cyc}2sqrt{a^3b^3cdot a^3c^2b}=2sum_{cyc}a^3b^2c.$$
Thus,
$$sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)geq10sum_{cyc}a^3b^2c.$$
Id est, it's enough to prove that
$$sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)geq24a^2b^2c^2,$$ which is AM-GM again.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
    $endgroup$
    – MathFanatics
    Jan 5 at 13:10












  • $begingroup$
    Which step? Rearrangement?
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:11










  • $begingroup$
    Yes that would be helpful
    $endgroup$
    – MathFanatics
    Jan 5 at 13:13










  • $begingroup$
    @MathFanatics I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:15










  • $begingroup$
    But will that work with the original inequalitys' coefficients?
    $endgroup$
    – MathFanatics
    Jan 5 at 13:17





















0












$begingroup$

Hint: Use AM-GM inequality, like:



$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3geq 24a^2b^2c^2$$
and then you are left to prove:



$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Should I use rearrangement on the rest of the terms?
    $endgroup$
    – MathFanatics
    Jan 5 at 12:47










  • $begingroup$
    I think AM-GMis enought
    $endgroup$
    – greedoid
    Jan 5 at 12:48










  • $begingroup$
    I tried to apply AM-GM on different pairs on the left hand side but it failed
    $endgroup$
    – MathFanatics
    Jan 5 at 12:50











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The hint:



You can use AM-GM and Rearrangement.



For example, by Rearrangement
$$sum_{cyc}(a^4bc-a^3b^2c)=abcsum_{cyc}(a^3-a^2b)geq0.$$
Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.



Thus, $$sum_{cyc}a^3=a^2cdot a+b^2cdot b+c^2cdot cgeq a^2cdot b+b^2cdot c+c^2cdot a=sum_{cyc}a^2b.$$



Now, by AM-GM
$$sum_{cyc}(a^3b^3+a^3c^2b)geqsum_{cyc}2sqrt{a^3b^3cdot a^3c^2b}=2sum_{cyc}a^3b^2c.$$
Thus,
$$sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)geq10sum_{cyc}a^3b^2c.$$
Id est, it's enough to prove that
$$sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)geq24a^2b^2c^2,$$ which is AM-GM again.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
    $endgroup$
    – MathFanatics
    Jan 5 at 13:10












  • $begingroup$
    Which step? Rearrangement?
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:11










  • $begingroup$
    Yes that would be helpful
    $endgroup$
    – MathFanatics
    Jan 5 at 13:13










  • $begingroup$
    @MathFanatics I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:15










  • $begingroup$
    But will that work with the original inequalitys' coefficients?
    $endgroup$
    – MathFanatics
    Jan 5 at 13:17


















0












$begingroup$

The hint:



You can use AM-GM and Rearrangement.



For example, by Rearrangement
$$sum_{cyc}(a^4bc-a^3b^2c)=abcsum_{cyc}(a^3-a^2b)geq0.$$
Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.



Thus, $$sum_{cyc}a^3=a^2cdot a+b^2cdot b+c^2cdot cgeq a^2cdot b+b^2cdot c+c^2cdot a=sum_{cyc}a^2b.$$



Now, by AM-GM
$$sum_{cyc}(a^3b^3+a^3c^2b)geqsum_{cyc}2sqrt{a^3b^3cdot a^3c^2b}=2sum_{cyc}a^3b^2c.$$
Thus,
$$sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)geq10sum_{cyc}a^3b^2c.$$
Id est, it's enough to prove that
$$sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)geq24a^2b^2c^2,$$ which is AM-GM again.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
    $endgroup$
    – MathFanatics
    Jan 5 at 13:10












  • $begingroup$
    Which step? Rearrangement?
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:11










  • $begingroup$
    Yes that would be helpful
    $endgroup$
    – MathFanatics
    Jan 5 at 13:13










  • $begingroup$
    @MathFanatics I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:15










  • $begingroup$
    But will that work with the original inequalitys' coefficients?
    $endgroup$
    – MathFanatics
    Jan 5 at 13:17
















0












0








0





$begingroup$

The hint:



You can use AM-GM and Rearrangement.



For example, by Rearrangement
$$sum_{cyc}(a^4bc-a^3b^2c)=abcsum_{cyc}(a^3-a^2b)geq0.$$
Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.



Thus, $$sum_{cyc}a^3=a^2cdot a+b^2cdot b+c^2cdot cgeq a^2cdot b+b^2cdot c+c^2cdot a=sum_{cyc}a^2b.$$



Now, by AM-GM
$$sum_{cyc}(a^3b^3+a^3c^2b)geqsum_{cyc}2sqrt{a^3b^3cdot a^3c^2b}=2sum_{cyc}a^3b^2c.$$
Thus,
$$sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)geq10sum_{cyc}a^3b^2c.$$
Id est, it's enough to prove that
$$sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)geq24a^2b^2c^2,$$ which is AM-GM again.






share|cite|improve this answer











$endgroup$



The hint:



You can use AM-GM and Rearrangement.



For example, by Rearrangement
$$sum_{cyc}(a^4bc-a^3b^2c)=abcsum_{cyc}(a^3-a^2b)geq0.$$
Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.



Thus, $$sum_{cyc}a^3=a^2cdot a+b^2cdot b+c^2cdot cgeq a^2cdot b+b^2cdot c+c^2cdot a=sum_{cyc}a^2b.$$



Now, by AM-GM
$$sum_{cyc}(a^3b^3+a^3c^2b)geqsum_{cyc}2sqrt{a^3b^3cdot a^3c^2b}=2sum_{cyc}a^3b^2c.$$
Thus,
$$sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)geq10sum_{cyc}a^3b^2c.$$
Id est, it's enough to prove that
$$sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)geq24a^2b^2c^2,$$ which is AM-GM again.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 13:20

























answered Jan 5 at 12:44









Michael RozenbergMichael Rozenberg

107k1894198




107k1894198












  • $begingroup$
    Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
    $endgroup$
    – MathFanatics
    Jan 5 at 13:10












  • $begingroup$
    Which step? Rearrangement?
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:11










  • $begingroup$
    Yes that would be helpful
    $endgroup$
    – MathFanatics
    Jan 5 at 13:13










  • $begingroup$
    @MathFanatics I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:15










  • $begingroup$
    But will that work with the original inequalitys' coefficients?
    $endgroup$
    – MathFanatics
    Jan 5 at 13:17




















  • $begingroup$
    Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
    $endgroup$
    – MathFanatics
    Jan 5 at 13:10












  • $begingroup$
    Which step? Rearrangement?
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:11










  • $begingroup$
    Yes that would be helpful
    $endgroup$
    – MathFanatics
    Jan 5 at 13:13










  • $begingroup$
    @MathFanatics I added something. See now.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:15










  • $begingroup$
    But will that work with the original inequalitys' coefficients?
    $endgroup$
    – MathFanatics
    Jan 5 at 13:17


















$begingroup$
Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
$endgroup$
– MathFanatics
Jan 5 at 13:10






$begingroup$
Could you explain the steps please? And I don't see how you've taken account the coefficients of the original inequality
$endgroup$
– MathFanatics
Jan 5 at 13:10














$begingroup$
Which step? Rearrangement?
$endgroup$
– Michael Rozenberg
Jan 5 at 13:11




$begingroup$
Which step? Rearrangement?
$endgroup$
– Michael Rozenberg
Jan 5 at 13:11












$begingroup$
Yes that would be helpful
$endgroup$
– MathFanatics
Jan 5 at 13:13




$begingroup$
Yes that would be helpful
$endgroup$
– MathFanatics
Jan 5 at 13:13












$begingroup$
@MathFanatics I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:15




$begingroup$
@MathFanatics I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:15












$begingroup$
But will that work with the original inequalitys' coefficients?
$endgroup$
– MathFanatics
Jan 5 at 13:17






$begingroup$
But will that work with the original inequalitys' coefficients?
$endgroup$
– MathFanatics
Jan 5 at 13:17













0












$begingroup$

Hint: Use AM-GM inequality, like:



$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3geq 24a^2b^2c^2$$
and then you are left to prove:



$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Should I use rearrangement on the rest of the terms?
    $endgroup$
    – MathFanatics
    Jan 5 at 12:47










  • $begingroup$
    I think AM-GMis enought
    $endgroup$
    – greedoid
    Jan 5 at 12:48










  • $begingroup$
    I tried to apply AM-GM on different pairs on the left hand side but it failed
    $endgroup$
    – MathFanatics
    Jan 5 at 12:50
















0












$begingroup$

Hint: Use AM-GM inequality, like:



$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3geq 24a^2b^2c^2$$
and then you are left to prove:



$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Should I use rearrangement on the rest of the terms?
    $endgroup$
    – MathFanatics
    Jan 5 at 12:47










  • $begingroup$
    I think AM-GMis enought
    $endgroup$
    – greedoid
    Jan 5 at 12:48










  • $begingroup$
    I tried to apply AM-GM on different pairs on the left hand side but it failed
    $endgroup$
    – MathFanatics
    Jan 5 at 12:50














0












0








0





$begingroup$

Hint: Use AM-GM inequality, like:



$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3geq 24a^2b^2c^2$$
and then you are left to prove:



$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$






share|cite|improve this answer











$endgroup$



Hint: Use AM-GM inequality, like:



$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3geq 24a^2b^2c^2$$
and then you are left to prove:



$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 12:48

























answered Jan 5 at 12:43









greedoidgreedoid

46k1160117




46k1160117












  • $begingroup$
    Should I use rearrangement on the rest of the terms?
    $endgroup$
    – MathFanatics
    Jan 5 at 12:47










  • $begingroup$
    I think AM-GMis enought
    $endgroup$
    – greedoid
    Jan 5 at 12:48










  • $begingroup$
    I tried to apply AM-GM on different pairs on the left hand side but it failed
    $endgroup$
    – MathFanatics
    Jan 5 at 12:50


















  • $begingroup$
    Should I use rearrangement on the rest of the terms?
    $endgroup$
    – MathFanatics
    Jan 5 at 12:47










  • $begingroup$
    I think AM-GMis enought
    $endgroup$
    – greedoid
    Jan 5 at 12:48










  • $begingroup$
    I tried to apply AM-GM on different pairs on the left hand side but it failed
    $endgroup$
    – MathFanatics
    Jan 5 at 12:50
















$begingroup$
Should I use rearrangement on the rest of the terms?
$endgroup$
– MathFanatics
Jan 5 at 12:47




$begingroup$
Should I use rearrangement on the rest of the terms?
$endgroup$
– MathFanatics
Jan 5 at 12:47












$begingroup$
I think AM-GMis enought
$endgroup$
– greedoid
Jan 5 at 12:48




$begingroup$
I think AM-GMis enought
$endgroup$
– greedoid
Jan 5 at 12:48












$begingroup$
I tried to apply AM-GM on different pairs on the left hand side but it failed
$endgroup$
– MathFanatics
Jan 5 at 12:50




$begingroup$
I tried to apply AM-GM on different pairs on the left hand side but it failed
$endgroup$
– MathFanatics
Jan 5 at 12:50


















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