$a,b,c>0$ and $abc=1$; prove $sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$ [duplicate] The...

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$a,b,c>0$ and $abc=1$; prove $sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$ [duplicate]



The Next CEO of Stack OverflowProve $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$Prove that $sumlimits_{cyc}frac{a}{b(3+a-b)}ge 1$For what values of $k>0$ does $abc=1 implies sum_{mbox{cyc}}left(frac1{a+k}-frac{a}{a^2+k}right) geq 0$?show $sum_{cyc}(1-x)^2ge sum_{cyc}frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$prove $sum_{cyc}frac{1}{a(a+b)}gefrac{4}{ac+bd}$Inequality $sum_{cyc} sqrt{frac{a}{a+8}} geq 1$prove this inequality $Σ_{cyc}frac{a^3+abc}{b^2+c^2}ge a+b+c$For $abc=1$ prove that $sumlimits_{cyc}frac{a}{a^{11}+1}leqfrac{3}{2}.$prove this inequality by $abc=1$If $ab+bc+ca=3$ for non-negative $a$, $b$, $c$, show that $sum_{cyc}a^2b^2+sum_{cyc}frac{12a^2b^2c^2}{(a+b)^2}ge 12abc$Prove the inequality $sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$ with Cauchy-Schwarz












0












$begingroup$



This question already has an answer here:




  • Prove $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$

    1 answer





Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$$




Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} ge displaystyle sum_{cyc}dfrac{1}{(c + 1)(a + 1)} = sum_{cyc}dfrac{1}{x - b + ca}$$
That means
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} + dfrac{1}{a + b + c + 1} ge dfrac{16}{bc + ca + ab + 3x + 1}$$
$$ge dfrac{48}{(x - 1)^2 + 9x + 3} = dfrac{48}{x^2 + 7x + 4}$$
That was illogical.






algebra-precalculus inequality substitution







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 Mar 17 at 12:06


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




    $begingroup$
    Title is very odd.
    $endgroup$
    – Love Invariants
    Mar 16 at 17:33
















0












$begingroup$



This question already has an answer here:




  • Prove $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$

    1 answer





Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$$




Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} ge displaystyle sum_{cyc}dfrac{1}{(c + 1)(a + 1)} = sum_{cyc}dfrac{1}{x - b + ca}$$
That means
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} + dfrac{1}{a + b + c + 1} ge dfrac{16}{bc + ca + ab + 3x + 1}$$
$$ge dfrac{48}{(x - 1)^2 + 9x + 3} = dfrac{48}{x^2 + 7x + 4}$$
That was illogical.






algebra-precalculus inequality substitution







share|cite|improve this question











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Users with the  inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.

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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




    $begingroup$
    Title is very odd.
    $endgroup$
    – Love Invariants
    Mar 16 at 17:33














0












0








0


1



$begingroup$



This question already has an answer here:




  • Prove $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$

    1 answer





Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$$




Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} ge displaystyle sum_{cyc}dfrac{1}{(c + 1)(a + 1)} = sum_{cyc}dfrac{1}{x - b + ca}$$
That means
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} + dfrac{1}{a + b + c + 1} ge dfrac{16}{bc + ca + ab + 3x + 1}$$
$$ge dfrac{48}{(x - 1)^2 + 9x + 3} = dfrac{48}{x^2 + 7x + 4}$$
That was illogical.






algebra-precalculus inequality substitution







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Prove $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$

    1 answer





Let $a$, $b$ and $c$ be three positives such that $abc=1$. Prove that $$sum_{cyc}frac1{(b+1)^2}+frac1{a+b+c+1}ge1$$




Here's what have I done that is completely incorrect. Let $a + b + c + 1 = x$. We have that
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} ge displaystyle sum_{cyc}dfrac{1}{(c + 1)(a + 1)} = sum_{cyc}dfrac{1}{x - b + ca}$$
That means
$$displaystyle sum_{cyc}dfrac{1}{(b + 1)^2} + dfrac{1}{a + b + c + 1} ge dfrac{16}{bc + ca + ab + 3x + 1}$$
$$ge dfrac{48}{(x - 1)^2 + 9x + 3} = dfrac{48}{x^2 + 7x + 4}$$
That was illogical.





This question already has an answer here:




  • Prove $Σ_{cyc}frac{1}{left(a+1right)^2}+frac{1}{a+b+c+1}ge 1$

    1 answer







algebra-precalculus inequality substitution




algebra-precalculus inequality substitution






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 5:15









Michael Rozenberg

109k1896201




109k1896201










asked Mar 16 at 17:13









Lê Thành ĐạtLê Thành Đạt

33312




33312




marked as duplicate by Carl Mummert, Michael Rozenberg inequality
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marked as duplicate by Carl Mummert, Michael Rozenberg inequality
Users with the  inequality badge can single-handedly close inequality questions as duplicates and reopen them as needed.

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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




    $begingroup$
    Title is very odd.
    $endgroup$
    – Love Invariants
    Mar 16 at 17:33














  • 1




    $begingroup$
    Title is very odd.
    $endgroup$
    – Love Invariants
    Mar 16 at 17:33








1




1




$begingroup$
Title is very odd.
$endgroup$
– Love Invariants
Mar 16 at 17:33




$begingroup$
Title is very odd.
$endgroup$
– Love Invariants
Mar 16 at 17:33










3 Answers
3






active

oldest

votes


















1












$begingroup$

Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.



Thus, $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}-1=$$
$$=sum_{cyc}frac{1}{(a+1)^2}+frac{2}{prodlimits_{cyc}(a+1)}-1+frac{1}{a+b+c+1}-frac{2}{prodlimits_{cyc}(a+1)}=$$
$$=frac{a^2+b^2+c^2-3}{prodlimits_{cyc}(a+1)^2}+frac{sumlimits_{cyc}(ab-a)}{(a+b+c+1)prodlimits_{cyc}(a+1)}.$$
Thus, we need to prove that
$$(a^2+b^2+c^2-3)(a+b+c+1)+sum_{cyc}(ab-a)prod_{cyc}(a+1)geq0$$ or
$$3v^4+9u^3w-6uv^2w-5uw^3-w^4geq0,$$ which is true because $ugeq vgeq w$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 0:44










  • $begingroup$
    @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 2:15










  • $begingroup$
    Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 2:58










  • $begingroup$
    @Lê Thành Đạt There is a very nice proof. But it's not mine.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 4:15










  • $begingroup$
    Please post the proof if it is not the other answer.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 4:18



















2












$begingroup$

Hint: Substituting $$a=frac{x}{y},b=frac{y}{z}c=frac{z}{x}$$
so we get



$${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3
,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3,{x}^{3}{y}^{
4}{z}^{2}-3,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{
2}-3,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y}
^{4}{z}^{4}+{y}^{3}{z}^{6}
geq 0$$
This is true by AM-GM.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    WA says the same thing: wolframalpha.com/input/…
    $endgroup$
    – Michael Rozenberg
    Mar 16 at 19:12










  • $begingroup$
    Should i misreaded the problem?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 16 at 19:13










  • $begingroup$
    I can't get around proving the inequality with this method. Do you have a different answer?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 11:27



















1












$begingroup$

Another solution.



Since $$prod_{cyc}(a-1)^2=prod_{cyc}((a-1)(b-1))geq0,$$ we can assume that $(a-1)(b-1)geq0,$ which gives
$$a+bleq ab+1=frac{1}{c}+1.$$



Also, we have $$frac{1}{(a+1)^2}+frac{1}{(b+1)^2}-frac{1}{ab+1}=frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}geq0$$
Id est,
$$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}geqfrac{1}{ab+1}+frac{1}{(c+1)^2}+frac{1}{ab+1+c+1}=$$
$$=frac{1}{frac{1}{c}+1}+frac{1}{(c+1)^2}+frac{1}{frac{1}{c}+1+c+1}=frac{c}{c+1}+frac{1}{(c+1)^2}+frac{c}{(c+1)^2}=1.$$






share|cite|improve this answer









$endgroup$




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.



    Thus, $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}-1=$$
    $$=sum_{cyc}frac{1}{(a+1)^2}+frac{2}{prodlimits_{cyc}(a+1)}-1+frac{1}{a+b+c+1}-frac{2}{prodlimits_{cyc}(a+1)}=$$
    $$=frac{a^2+b^2+c^2-3}{prodlimits_{cyc}(a+1)^2}+frac{sumlimits_{cyc}(ab-a)}{(a+b+c+1)prodlimits_{cyc}(a+1)}.$$
    Thus, we need to prove that
    $$(a^2+b^2+c^2-3)(a+b+c+1)+sum_{cyc}(ab-a)prod_{cyc}(a+1)geq0$$ or
    $$3v^4+9u^3w-6uv^2w-5uw^3-w^4geq0,$$ which is true because $ugeq vgeq w$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 0:44










    • $begingroup$
      @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 2:15










    • $begingroup$
      Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 2:58










    • $begingroup$
      @Lê Thành Đạt There is a very nice proof. But it's not mine.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 4:15










    • $begingroup$
      Please post the proof if it is not the other answer.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 4:18
















    1












    $begingroup$

    Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.



    Thus, $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}-1=$$
    $$=sum_{cyc}frac{1}{(a+1)^2}+frac{2}{prodlimits_{cyc}(a+1)}-1+frac{1}{a+b+c+1}-frac{2}{prodlimits_{cyc}(a+1)}=$$
    $$=frac{a^2+b^2+c^2-3}{prodlimits_{cyc}(a+1)^2}+frac{sumlimits_{cyc}(ab-a)}{(a+b+c+1)prodlimits_{cyc}(a+1)}.$$
    Thus, we need to prove that
    $$(a^2+b^2+c^2-3)(a+b+c+1)+sum_{cyc}(ab-a)prod_{cyc}(a+1)geq0$$ or
    $$3v^4+9u^3w-6uv^2w-5uw^3-w^4geq0,$$ which is true because $ugeq vgeq w$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 0:44










    • $begingroup$
      @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 2:15










    • $begingroup$
      Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 2:58










    • $begingroup$
      @Lê Thành Đạt There is a very nice proof. But it's not mine.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 4:15










    • $begingroup$
      Please post the proof if it is not the other answer.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 4:18














    1












    1








    1





    $begingroup$

    Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.



    Thus, $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}-1=$$
    $$=sum_{cyc}frac{1}{(a+1)^2}+frac{2}{prodlimits_{cyc}(a+1)}-1+frac{1}{a+b+c+1}-frac{2}{prodlimits_{cyc}(a+1)}=$$
    $$=frac{a^2+b^2+c^2-3}{prodlimits_{cyc}(a+1)^2}+frac{sumlimits_{cyc}(ab-a)}{(a+b+c+1)prodlimits_{cyc}(a+1)}.$$
    Thus, we need to prove that
    $$(a^2+b^2+c^2-3)(a+b+c+1)+sum_{cyc}(ab-a)prod_{cyc}(a+1)geq0$$ or
    $$3v^4+9u^3w-6uv^2w-5uw^3-w^4geq0,$$ which is true because $ugeq vgeq w$.






    share|cite|improve this answer









    $endgroup$



    Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.



    Thus, $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}-1=$$
    $$=sum_{cyc}frac{1}{(a+1)^2}+frac{2}{prodlimits_{cyc}(a+1)}-1+frac{1}{a+b+c+1}-frac{2}{prodlimits_{cyc}(a+1)}=$$
    $$=frac{a^2+b^2+c^2-3}{prodlimits_{cyc}(a+1)^2}+frac{sumlimits_{cyc}(ab-a)}{(a+b+c+1)prodlimits_{cyc}(a+1)}.$$
    Thus, we need to prove that
    $$(a^2+b^2+c^2-3)(a+b+c+1)+sum_{cyc}(ab-a)prod_{cyc}(a+1)geq0$$ or
    $$3v^4+9u^3w-6uv^2w-5uw^3-w^4geq0,$$ which is true because $ugeq vgeq w$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 16 at 18:17









    Michael RozenbergMichael Rozenberg

    109k1896201




    109k1896201












    • $begingroup$
      Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 0:44










    • $begingroup$
      @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 2:15










    • $begingroup$
      Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 2:58










    • $begingroup$
      @Lê Thành Đạt There is a very nice proof. But it's not mine.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 4:15










    • $begingroup$
      Please post the proof if it is not the other answer.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 4:18


















    • $begingroup$
      Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 0:44










    • $begingroup$
      @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 2:15










    • $begingroup$
      Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 2:58










    • $begingroup$
      @Lê Thành Đạt There is a very nice proof. But it's not mine.
      $endgroup$
      – Michael Rozenberg
      Mar 17 at 4:15










    • $begingroup$
      Please post the proof if it is not the other answer.
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 4:18
















    $begingroup$
    Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 0:44




    $begingroup$
    Could you fix your answer if possible because the given condition is already $abc = 1$. There's no need to let $abc = w^3$.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 0:44












    $begingroup$
    @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 2:15




    $begingroup$
    @Lê Thành Đạt Just $w=1$, which gives a homogenization. See please better my solution.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 2:15












    $begingroup$
    Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 2:58




    $begingroup$
    Is there a way to prove the inequation without using the $uvw$'s method and by using classic inequalities?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 2:58












    $begingroup$
    @Lê Thành Đạt There is a very nice proof. But it's not mine.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 4:15




    $begingroup$
    @Lê Thành Đạt There is a very nice proof. But it's not mine.
    $endgroup$
    – Michael Rozenberg
    Mar 17 at 4:15












    $begingroup$
    Please post the proof if it is not the other answer.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 4:18




    $begingroup$
    Please post the proof if it is not the other answer.
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 4:18











    2












    $begingroup$

    Hint: Substituting $$a=frac{x}{y},b=frac{y}{z}c=frac{z}{x}$$
    so we get



    $${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3
    ,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3,{x}^{3}{y}^{
    4}{z}^{2}-3,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{
    2}-3,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y}
    ^{4}{z}^{4}+{y}^{3}{z}^{6}
    geq 0$$
    This is true by AM-GM.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      WA says the same thing: wolframalpha.com/input/…
      $endgroup$
      – Michael Rozenberg
      Mar 16 at 19:12










    • $begingroup$
      Should i misreaded the problem?
      $endgroup$
      – Dr. Sonnhard Graubner
      Mar 16 at 19:13










    • $begingroup$
      I can't get around proving the inequality with this method. Do you have a different answer?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 11:27
















    2












    $begingroup$

    Hint: Substituting $$a=frac{x}{y},b=frac{y}{z}c=frac{z}{x}$$
    so we get



    $${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3
    ,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3,{x}^{3}{y}^{
    4}{z}^{2}-3,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{
    2}-3,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y}
    ^{4}{z}^{4}+{y}^{3}{z}^{6}
    geq 0$$
    This is true by AM-GM.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      WA says the same thing: wolframalpha.com/input/…
      $endgroup$
      – Michael Rozenberg
      Mar 16 at 19:12










    • $begingroup$
      Should i misreaded the problem?
      $endgroup$
      – Dr. Sonnhard Graubner
      Mar 16 at 19:13










    • $begingroup$
      I can't get around proving the inequality with this method. Do you have a different answer?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 11:27














    2












    2








    2





    $begingroup$

    Hint: Substituting $$a=frac{x}{y},b=frac{y}{z}c=frac{z}{x}$$
    so we get



    $${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3
    ,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3,{x}^{3}{y}^{
    4}{z}^{2}-3,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{
    2}-3,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y}
    ^{4}{z}^{4}+{y}^{3}{z}^{6}
    geq 0$$
    This is true by AM-GM.






    share|cite|improve this answer









    $endgroup$



    Hint: Substituting $$a=frac{x}{y},b=frac{y}{z}c=frac{z}{x}$$
    so we get



    $${x}^{6}{z}^{3}+{x}^{5}{y}^{3}z+{x}^{5}{y}^{2}{z}^{2}+{x}^{4}{y}^{4}z-3
    ,{x}^{4}{y}^{2}{z}^{3}+{x}^{4}y{z}^{4}+{x}^{3}{y}^{6}-3,{x}^{3}{y}^{
    4}{z}^{2}-3,{x}^{3}{y}^{3}{z}^{3}+{x}^{3}y{z}^{5}+{x}^{2}{y}^{5}{z}^{
    2}-3,{x}^{2}{y}^{3}{z}^{4}+{x}^{2}{y}^{2}{z}^{5}+x{y}^{5}{z}^{3}+x{y}
    ^{4}{z}^{4}+{y}^{3}{z}^{6}
    geq 0$$
    This is true by AM-GM.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 16 at 17:35









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    78.4k42867




    78.4k42867












    • $begingroup$
      WA says the same thing: wolframalpha.com/input/…
      $endgroup$
      – Michael Rozenberg
      Mar 16 at 19:12










    • $begingroup$
      Should i misreaded the problem?
      $endgroup$
      – Dr. Sonnhard Graubner
      Mar 16 at 19:13










    • $begingroup$
      I can't get around proving the inequality with this method. Do you have a different answer?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 11:27


















    • $begingroup$
      WA says the same thing: wolframalpha.com/input/…
      $endgroup$
      – Michael Rozenberg
      Mar 16 at 19:12










    • $begingroup$
      Should i misreaded the problem?
      $endgroup$
      – Dr. Sonnhard Graubner
      Mar 16 at 19:13










    • $begingroup$
      I can't get around proving the inequality with this method. Do you have a different answer?
      $endgroup$
      – Lê Thành Đạt
      Mar 17 at 11:27
















    $begingroup$
    WA says the same thing: wolframalpha.com/input/…
    $endgroup$
    – Michael Rozenberg
    Mar 16 at 19:12




    $begingroup$
    WA says the same thing: wolframalpha.com/input/…
    $endgroup$
    – Michael Rozenberg
    Mar 16 at 19:12












    $begingroup$
    Should i misreaded the problem?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 16 at 19:13




    $begingroup$
    Should i misreaded the problem?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 16 at 19:13












    $begingroup$
    I can't get around proving the inequality with this method. Do you have a different answer?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 11:27




    $begingroup$
    I can't get around proving the inequality with this method. Do you have a different answer?
    $endgroup$
    – Lê Thành Đạt
    Mar 17 at 11:27











    1












    $begingroup$

    Another solution.



    Since $$prod_{cyc}(a-1)^2=prod_{cyc}((a-1)(b-1))geq0,$$ we can assume that $(a-1)(b-1)geq0,$ which gives
    $$a+bleq ab+1=frac{1}{c}+1.$$



    Also, we have $$frac{1}{(a+1)^2}+frac{1}{(b+1)^2}-frac{1}{ab+1}=frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}geq0$$
    Id est,
    $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}geqfrac{1}{ab+1}+frac{1}{(c+1)^2}+frac{1}{ab+1+c+1}=$$
    $$=frac{1}{frac{1}{c}+1}+frac{1}{(c+1)^2}+frac{1}{frac{1}{c}+1+c+1}=frac{c}{c+1}+frac{1}{(c+1)^2}+frac{c}{(c+1)^2}=1.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Another solution.



      Since $$prod_{cyc}(a-1)^2=prod_{cyc}((a-1)(b-1))geq0,$$ we can assume that $(a-1)(b-1)geq0,$ which gives
      $$a+bleq ab+1=frac{1}{c}+1.$$



      Also, we have $$frac{1}{(a+1)^2}+frac{1}{(b+1)^2}-frac{1}{ab+1}=frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}geq0$$
      Id est,
      $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}geqfrac{1}{ab+1}+frac{1}{(c+1)^2}+frac{1}{ab+1+c+1}=$$
      $$=frac{1}{frac{1}{c}+1}+frac{1}{(c+1)^2}+frac{1}{frac{1}{c}+1+c+1}=frac{c}{c+1}+frac{1}{(c+1)^2}+frac{c}{(c+1)^2}=1.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Another solution.



        Since $$prod_{cyc}(a-1)^2=prod_{cyc}((a-1)(b-1))geq0,$$ we can assume that $(a-1)(b-1)geq0,$ which gives
        $$a+bleq ab+1=frac{1}{c}+1.$$



        Also, we have $$frac{1}{(a+1)^2}+frac{1}{(b+1)^2}-frac{1}{ab+1}=frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}geq0$$
        Id est,
        $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}geqfrac{1}{ab+1}+frac{1}{(c+1)^2}+frac{1}{ab+1+c+1}=$$
        $$=frac{1}{frac{1}{c}+1}+frac{1}{(c+1)^2}+frac{1}{frac{1}{c}+1+c+1}=frac{c}{c+1}+frac{1}{(c+1)^2}+frac{c}{(c+1)^2}=1.$$






        share|cite|improve this answer









        $endgroup$



        Another solution.



        Since $$prod_{cyc}(a-1)^2=prod_{cyc}((a-1)(b-1))geq0,$$ we can assume that $(a-1)(b-1)geq0,$ which gives
        $$a+bleq ab+1=frac{1}{c}+1.$$



        Also, we have $$frac{1}{(a+1)^2}+frac{1}{(b+1)^2}-frac{1}{ab+1}=frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2}geq0$$
        Id est,
        $$sum_{cyc}frac{1}{(a+1)^2}+frac{1}{a+b+c+1}geqfrac{1}{ab+1}+frac{1}{(c+1)^2}+frac{1}{ab+1+c+1}=$$
        $$=frac{1}{frac{1}{c}+1}+frac{1}{(c+1)^2}+frac{1}{frac{1}{c}+1+c+1}=frac{c}{c+1}+frac{1}{(c+1)^2}+frac{c}{(c+1)^2}=1.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 17 at 5:03









        Michael RozenbergMichael Rozenberg

        109k1896201




        109k1896201















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