Prove that $sqrtfraca^2+b^2+c^2+d^24 ge sqrt[3]fracabc+bcd+cda+dab4$. [closed]

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Let $a$, $b$, $c$ and $d$ be some positive integers. Prove that:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[3]fracabc+bcd+cda+dab4$$



I was trying to use $A.M. ge G.M.$, but it's not enough:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[4]abcd$$



Can you help me, please? Thanks!



Edit: I have just now seen that this question is a duplicate. I didn't saw it in the list of suggested questions when I was writing it. Sorry.




inequality integral-inequality




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closed as off-topic by Carl Mummert, Mostafa Ayaz, amWhy, Isaac Browne, Aaron Montgomery Jul 30 at 23:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Mostafa Ayaz, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
    – Carl Mummert
    Jul 30 at 19:26






  • 2




    I think this question does not miss context and in this question just there are details.
    – Michael Rozenberg
    Jul 30 at 19:32







  • 2




    Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
    – Isaac Browne
    Jul 30 at 21:06










  • I have already said that in the edit from the post.
    – Iulian Oleniuc
    Jul 30 at 21:09














up vote
0
down vote

favorite












Let $a$, $b$, $c$ and $d$ be some positive integers. Prove that:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[3]fracabc+bcd+cda+dab4$$



I was trying to use $A.M. ge G.M.$, but it's not enough:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[4]abcd$$



Can you help me, please? Thanks!



Edit: I have just now seen that this question is a duplicate. I didn't saw it in the list of suggested questions when I was writing it. Sorry.




inequality integral-inequality




share|cite|improve this question













closed as off-topic by Carl Mummert, Mostafa Ayaz, amWhy, Isaac Browne, Aaron Montgomery Jul 30 at 23:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Mostafa Ayaz, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
    – Carl Mummert
    Jul 30 at 19:26






  • 2




    I think this question does not miss context and in this question just there are details.
    – Michael Rozenberg
    Jul 30 at 19:32







  • 2




    Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
    – Isaac Browne
    Jul 30 at 21:06










  • I have already said that in the edit from the post.
    – Iulian Oleniuc
    Jul 30 at 21:09












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $a$, $b$, $c$ and $d$ be some positive integers. Prove that:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[3]fracabc+bcd+cda+dab4$$



I was trying to use $A.M. ge G.M.$, but it's not enough:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[4]abcd$$



Can you help me, please? Thanks!



Edit: I have just now seen that this question is a duplicate. I didn't saw it in the list of suggested questions when I was writing it. Sorry.




inequality integral-inequality




share|cite|improve this question













Let $a$, $b$, $c$ and $d$ be some positive integers. Prove that:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[3]fracabc+bcd+cda+dab4$$



I was trying to use $A.M. ge G.M.$, but it's not enough:



$$sqrtfraca^2+b^2+c^2+d^24 ge sqrt[4]abcd$$



Can you help me, please? Thanks!



Edit: I have just now seen that this question is a duplicate. I didn't saw it in the list of suggested questions when I was writing it. Sorry.





inequality integral-inequality





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 20:05
























asked Jul 30 at 19:24









Iulian Oleniuc

3619




3619




closed as off-topic by Carl Mummert, Mostafa Ayaz, amWhy, Isaac Browne, Aaron Montgomery Jul 30 at 23:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Mostafa Ayaz, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Carl Mummert, Mostafa Ayaz, amWhy, Isaac Browne, Aaron Montgomery Jul 30 at 23:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Mostafa Ayaz, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
    – Carl Mummert
    Jul 30 at 19:26






  • 2




    I think this question does not miss context and in this question just there are details.
    – Michael Rozenberg
    Jul 30 at 19:32







  • 2




    Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
    – Isaac Browne
    Jul 30 at 21:06










  • I have already said that in the edit from the post.
    – Iulian Oleniuc
    Jul 30 at 21:09












  • 1




    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
    – Carl Mummert
    Jul 30 at 19:26






  • 2




    I think this question does not miss context and in this question just there are details.
    – Michael Rozenberg
    Jul 30 at 19:32







  • 2




    Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
    – Isaac Browne
    Jul 30 at 21:06










  • I have already said that in the edit from the post.
    – Iulian Oleniuc
    Jul 30 at 21:09







1




1




This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
– Carl Mummert
Jul 30 at 19:26




This question is missing context or other details: Please improve the question by providing additional context, which ideally includes the source and motivation of the problem, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
– Carl Mummert
Jul 30 at 19:26




2




2




I think this question does not miss context and in this question just there are details.
– Michael Rozenberg
Jul 30 at 19:32





I think this question does not miss context and in this question just there are details.
– Michael Rozenberg
Jul 30 at 19:32





2




2




Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
– Isaac Browne
Jul 30 at 21:06




Possible duplicate of How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$
– Isaac Browne
Jul 30 at 21:06












I have already said that in the edit from the post.
– Iulian Oleniuc
Jul 30 at 21:09




I have already said that in the edit from the post.
– Iulian Oleniuc
Jul 30 at 21:09










1 Answer
1






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oldest

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0
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accepted










By AM-GM and Maclaurin we obtain:
$$sqrtfraca^2+b^2+c^2+d^24geqsqrtfracab+ac+ad+bc+bd+cd6geqsqrt[3]fracabc+abd+acd+bcd4.$$
A proof of the second inequality see here:



How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$






share|cite|improve this answer





















  • Why down voted?
    – Michael Rozenberg
    Jul 30 at 21:05

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










By AM-GM and Maclaurin we obtain:
$$sqrtfraca^2+b^2+c^2+d^24geqsqrtfracab+ac+ad+bc+bd+cd6geqsqrt[3]fracabc+abd+acd+bcd4.$$
A proof of the second inequality see here:



How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$






share|cite|improve this answer





















  • Why down voted?
    – Michael Rozenberg
    Jul 30 at 21:05














up vote
0
down vote



accepted










By AM-GM and Maclaurin we obtain:
$$sqrtfraca^2+b^2+c^2+d^24geqsqrtfracab+ac+ad+bc+bd+cd6geqsqrt[3]fracabc+abd+acd+bcd4.$$
A proof of the second inequality see here:



How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$






share|cite|improve this answer





















  • Why down voted?
    – Michael Rozenberg
    Jul 30 at 21:05












up vote
0
down vote



accepted







up vote
0
down vote



accepted






By AM-GM and Maclaurin we obtain:
$$sqrtfraca^2+b^2+c^2+d^24geqsqrtfracab+ac+ad+bc+bd+cd6geqsqrt[3]fracabc+abd+acd+bcd4.$$
A proof of the second inequality see here:



How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$






share|cite|improve this answer













By AM-GM and Maclaurin we obtain:
$$sqrtfraca^2+b^2+c^2+d^24geqsqrtfracab+ac+ad+bc+bd+cd6geqsqrt[3]fracabc+abd+acd+bcd4.$$
A proof of the second inequality see here:



How to prove this inequality $sqrtfracab+bc+cd+da+ac+bd6geq sqrt[3]fracabc+bcd+cda+dab4$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 30 at 19:31









Michael Rozenberg

87.7k1578179




87.7k1578179











  • Why down voted?
    – Michael Rozenberg
    Jul 30 at 21:05
















  • Why down voted?
    – Michael Rozenberg
    Jul 30 at 21:05















Why down voted?
– Michael Rozenberg
Jul 30 at 21:05




Why down voted?
– Michael Rozenberg
Jul 30 at 21:05


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