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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062675%2falgebraic-inequality-for-positive-reals-a-b-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062675%2falgebraic-inequality-for-positive-reals-a-b-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more