Prove that $sum_cycfraca^2ca^2 + 2c^2 ge 1$ [duplicate] The 2019 Stack Overflow Developer Survey Results Are InHow prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$How prove this inequality $sumlimits_cycfraca^3b+c+dge dfrac13$,if $sumlimits_cycasqrtbcge 1$If $x+y+z=3$, then $sum_textcycfracx^22y^2-y+3gefrac34$Prove $sumlimits_textcycfracaa+(n-1)bgeq 1$To prove $sum_cycfrac1a^3+b^3+abc le frac1abc$If $a+b+c = 3$ show $9 + 3 sum_mboxcycacosleft( frac2bcright)geq 2left( sum_mboxcycacosleft( fracbcright) right)^2$show $sum_cyc(1-x)^2ge sum_cycfracz^2(1-x^2)(1-y^2)(xy+z)^2.$Inequality : $sum_cycfracsqrta^3c2sqrtb^3a+3bcgeq frac35$If $ab+bc+ca=3$ for non-negative $a$, $b$, $c$, show that $sum_cyca^2b^2+sum_cycfrac12a^2b^2c^2(a+b)^2ge 12abc$$a,b,c>0$ and $abc=1$; prove $sum_cycfrac1(b+1)^2+frac1a+b+c+1ge1$Prove that $sum_cycdfracaa + b^4 + c^4 le 1$ where $abc = 1$.

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Prove that $sum_cycfraca^2ca^2 + 2c^2 ge 1$ [duplicate]



The 2019 Stack Overflow Developer Survey Results Are InHow prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$How prove this inequality $sumlimits_cycfraca^3b+c+dge dfrac13$,if $sumlimits_cycasqrtbcge 1$If $x+y+z=3$, then $sum_textcycfracx^22y^2-y+3gefrac34$Prove $sumlimits_textcycfracaa+(n-1)bgeq 1$To prove $sum_cycfrac1a^3+b^3+abc le frac1abc$If $a+b+c = 3$ show $9 + 3 sum_mboxcycacosleft( frac2bcright)geq 2left( sum_mboxcycacosleft( fracbcright) right)^2$show $sum_cyc(1-x)^2ge sum_cycfracz^2(1-x^2)(1-y^2)(xy+z)^2.$Inequality : $sum_cycfracsqrta^3c2sqrtb^3a+3bcgeq frac35$If $ab+bc+ca=3$ for non-negative $a$, $b$, $c$, show that $sum_cyca^2b^2+sum_cycfrac12a^2b^2c^2(a+b)^2ge 12abc$$a,b,c>0$ and $abc=1$; prove $sum_cycfrac1(b+1)^2+frac1a+b+c+1ge1$Prove that $sum_cycdfracaa + b^4 + c^4 le 1$ where $abc = 1$.










0












$begingroup$



This question already has an answer here:



  • How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$

    1 answer




$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$




Here's what I did. My stupidity has reached a spiritual level.



We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.



$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$



$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$



$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$



$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$



I am done with my life.










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$endgroup$



marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26


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




    $begingroup$
    is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 23 at 17:34






  • 6




    $begingroup$
    Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
    $endgroup$
    – Word Shallow
    Mar 24 at 2:46
















0












$begingroup$



This question already has an answer here:



  • How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$

    1 answer




$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$




Here's what I did. My stupidity has reached a spiritual level.



We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.



$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$



$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$



$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$



$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$



I am done with my life.










share|cite|improve this question











$endgroup$



marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26


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




    $begingroup$
    is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 23 at 17:34






  • 6




    $begingroup$
    Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
    $endgroup$
    – Word Shallow
    Mar 24 at 2:46














0












0








0


3



$begingroup$



This question already has an answer here:



  • How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$

    1 answer




$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$




Here's what I did. My stupidity has reached a spiritual level.



We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.



$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$



$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$



$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$



$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$



I am done with my life.










share|cite|improve this question











$endgroup$





This question already has an answer here:



  • How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$

    1 answer




$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ sum_cycfraca^2ca^2 + 2c^2 ge 1$$




Here's what I did. My stupidity has reached a spiritual level.



We have that $ab + bc + ca = 3abc implies dfrac1a + dfrac1b + dfrac1c = 3$.



$$sum_cycdfraca^2ca^2 + 2c^2 = sum_cycdfrac1cleft(1 - dfrac2ca^2 + 2cright) ge sum_cycdfrac1cleft(1 - dfrac2c2c + 2a - 1right)$$



$$ = left(dfrac1a + dfrac1b + dfrac1cright) - 2sum_cycdfrac12c + 2a - 1 ge 3 - dfrac29sum_cycleft(dfrac1c + dfrac1a + dfrac1c + a - 1right)$$



$$ = 3 - dfrac49left(dfrac1a + dfrac1b + dfrac1cright) - dfrac29sum_cycdfrac1c + a - 1 ge 3 - dfrac43 - dfrac118sum_cycleft(dfrac1c - frac12 + dfrac1a - frac12right)$$



$$ = dfrac53 - dfrac29left(dfrac12a - 1 + dfrac12b - 1 + dfrac12c -1right)$$



I am done with my life.





This question already has an answer here:



  • How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$

    1 answer







inequality substitution cauchy-schwarz-inequality






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 18:08









StubbornAtom

6,37831440




6,37831440










asked Mar 23 at 17:29









Lê Thành ĐạtLê Thành Đạt

47313




47313




marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26


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 Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFan Mar 24 at 13:26


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




    $begingroup$
    is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 23 at 17:34






  • 6




    $begingroup$
    Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
    $endgroup$
    – Word Shallow
    Mar 24 at 2:46













  • 1




    $begingroup$
    is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 23 at 17:34






  • 6




    $begingroup$
    Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
    $endgroup$
    – Word Shallow
    Mar 24 at 2:46








1




1




$begingroup$
is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:34




$begingroup$
is this $$fraca^2ca^2+2c^2+fracb^2ab^2+2a^2+fracc^2bc^2+2b^2geq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:34




6




6




$begingroup$
Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
$endgroup$
– Word Shallow
Mar 24 at 2:46





$begingroup$
Let $(1/a;1/b;1/c)->(x;y;z)$ Possible duplicate of How prove $fraca^2a+2b^2+fracb^2b+2c^2+fracc^2c+2a^2ge 1$
$endgroup$
– Word Shallow
Mar 24 at 2:46











1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.



Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Power means works, right?
    $endgroup$
    – user574848
    Mar 24 at 7:37










  • $begingroup$
    @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
    $endgroup$
    – Michael Rozenberg
    Mar 24 at 7:39











  • $begingroup$
    oh I misread xyz=1
    $endgroup$
    – user574848
    Mar 24 at 7:41

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.



Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Power means works, right?
    $endgroup$
    – user574848
    Mar 24 at 7:37










  • $begingroup$
    @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
    $endgroup$
    – Michael Rozenberg
    Mar 24 at 7:39











  • $begingroup$
    oh I misread xyz=1
    $endgroup$
    – user574848
    Mar 24 at 7:41















1












$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.



Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Power means works, right?
    $endgroup$
    – user574848
    Mar 24 at 7:37










  • $begingroup$
    @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
    $endgroup$
    – Michael Rozenberg
    Mar 24 at 7:39











  • $begingroup$
    oh I misread xyz=1
    $endgroup$
    – user574848
    Mar 24 at 7:41













1












1








1





$begingroup$

Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.



Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?






share|cite|improve this answer









$endgroup$



Let $frac1a=x$, $frac1b=y$ and $frac1c=z$.



Thus, $x+y+z=3$ and by C-S we obtain:
$$sum_cycfraca^2ca^2+2c^2=sum_cycfracz^22x^2+z=sum_cycfracz^42x^2z^2+z^3geqfrac(x^2+y^2+z^2)^2sumlimits_cyc(2x^2y^2+x^3).$$
Id est, it's enough to prove that
$$(x^2+y^2+z^2)^2geqsumlimits_cyc(2x^2y^2+x^3)$$ or
$$x^4+y^4+z^4geq x^3+y^3+z^3.$$
Can you end it now?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 23 at 17:53









Michael RozenbergMichael Rozenberg

110k1896201




110k1896201











  • $begingroup$
    Power means works, right?
    $endgroup$
    – user574848
    Mar 24 at 7:37










  • $begingroup$
    @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
    $endgroup$
    – Michael Rozenberg
    Mar 24 at 7:39











  • $begingroup$
    oh I misread xyz=1
    $endgroup$
    – user574848
    Mar 24 at 7:41
















  • $begingroup$
    Power means works, right?
    $endgroup$
    – user574848
    Mar 24 at 7:37










  • $begingroup$
    @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
    $endgroup$
    – Michael Rozenberg
    Mar 24 at 7:39











  • $begingroup$
    oh I misread xyz=1
    $endgroup$
    – user574848
    Mar 24 at 7:41















$begingroup$
Power means works, right?
$endgroup$
– user574848
Mar 24 at 7:37




$begingroup$
Power means works, right?
$endgroup$
– user574848
Mar 24 at 7:37












$begingroup$
@user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
$endgroup$
– Michael Rozenberg
Mar 24 at 7:39





$begingroup$
@user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $sumlimits_cyc(x^4-x^3)=sumlimits_cyc(x^4-x^3-x+1)=sumlimits_cyc(x-1)^2(x^2+x+1)geq0.$
$endgroup$
– Michael Rozenberg
Mar 24 at 7:39













$begingroup$
oh I misread xyz=1
$endgroup$
– user574848
Mar 24 at 7:41




$begingroup$
oh I misread xyz=1
$endgroup$
– user574848
Mar 24 at 7:41



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