Prove that $({aover a+b})^3+({bover b+c})^3+ ({cover c+a})^3geq {3over 8}$












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Let $a,b,c$ be positive real numbers. Prove that $$Big({aover a+b}Big)^3+Big({bover b+c}Big)^3+ Big({cover c+a}Big)^3geq {3over 8}$$




If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and



$$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq {3over 8}$$



Since $f(x)=Big({1over 1+x}Big)^3$ is convex we get, by Jensen,: $$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq 3f({x+y+zover 3})$$



Unfortunately, since $f$ is decreasing we don't have $f({x+y+zover 3}) geq f(1) = {1over 8}$.



Some idea how to solve this?










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  • 4




    $begingroup$
    How is this question "seeking personal advice," whoever voted to close this thread?
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:26


















11












$begingroup$



Let $a,b,c$ be positive real numbers. Prove that $$Big({aover a+b}Big)^3+Big({bover b+c}Big)^3+ Big({cover c+a}Big)^3geq {3over 8}$$




If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and



$$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq {3over 8}$$



Since $f(x)=Big({1over 1+x}Big)^3$ is convex we get, by Jensen,: $$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq 3f({x+y+zover 3})$$



Unfortunately, since $f$ is decreasing we don't have $f({x+y+zover 3}) geq f(1) = {1over 8}$.



Some idea how to solve this?










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    How is this question "seeking personal advice," whoever voted to close this thread?
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:26
















11












11








11


4



$begingroup$



Let $a,b,c$ be positive real numbers. Prove that $$Big({aover a+b}Big)^3+Big({bover b+c}Big)^3+ Big({cover c+a}Big)^3geq {3over 8}$$




If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and



$$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq {3over 8}$$



Since $f(x)=Big({1over 1+x}Big)^3$ is convex we get, by Jensen,: $$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq 3f({x+y+zover 3})$$



Unfortunately, since $f$ is decreasing we don't have $f({x+y+zover 3}) geq f(1) = {1over 8}$.



Some idea how to solve this?










share|cite|improve this question











$endgroup$





Let $a,b,c$ be positive real numbers. Prove that $$Big({aover a+b}Big)^3+Big({bover b+c}Big)^3+ Big({cover c+a}Big)^3geq {3over 8}$$




If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and



$$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq {3over 8}$$



Since $f(x)=Big({1over 1+x}Big)^3$ is convex we get, by Jensen,: $$Big({1over 1+x}Big)^3+Big({1over 1+y}Big)^3+ Big({1over 1+z}Big)^3geq 3f({x+y+zover 3})$$



Unfortunately, since $f$ is decreasing we don't have $f({x+y+zover 3}) geq f(1) = {1over 8}$.



Some idea how to solve this?







inequality cauchy-schwarz-inequality holder-inequality jensen-inequality uvw






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edited Dec 11 '18 at 19:51









Xander Henderson

14.3k103554




14.3k103554










asked Dec 4 '18 at 16:51









greedoidgreedoid

40.7k1149100




40.7k1149100








  • 4




    $begingroup$
    How is this question "seeking personal advice," whoever voted to close this thread?
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:26
















  • 4




    $begingroup$
    How is this question "seeking personal advice," whoever voted to close this thread?
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:26










4




4




$begingroup$
How is this question "seeking personal advice," whoever voted to close this thread?
$endgroup$
– Batominovski
Dec 4 '18 at 21:26






$begingroup$
How is this question "seeking personal advice," whoever voted to close this thread?
$endgroup$
– Batominovski
Dec 4 '18 at 21:26












3 Answers
3






active

oldest

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4












$begingroup$

Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
Thus, $g$ is convex on $big[-ln(3),inftybig)$.



Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
$$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
$$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
$$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't get it. What is the difference with my solution?
    $endgroup$
    – greedoid
    Dec 4 '18 at 21:45










  • $begingroup$
    I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:48





















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By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
Thus, it's enough to prove that
$$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
$$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
Now, by C-S
$$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it's enough to prove that
$$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
which is true even for all reals $a$, $b$ and $c$.



Indeed, the last inequality is symmetric inequality by degree four,



which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )



it's enough to prove the last inequality for equality case of two variables and since



it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
$$(a-1)^2(a+3)^2geq0.$$
Done!






share|cite|improve this answer









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    5












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    This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.



    Added:



    We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$



    The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.






    share|cite|improve this answer











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    • 2




      $begingroup$
      It seems to me there's enough here to post as an answer, no need to apologize.
      $endgroup$
      – David K
      Dec 4 '18 at 21:11











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    3 Answers
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    3 Answers
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    active

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    4












    $begingroup$

    Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
    Thus, $g$ is convex on $big[-ln(3),inftybig)$.



    Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
    If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
    $$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
    If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
    Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't get it. What is the difference with my solution?
      $endgroup$
      – greedoid
      Dec 4 '18 at 21:45










    • $begingroup$
      I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
      $endgroup$
      – Batominovski
      Dec 4 '18 at 21:48


















    4












    $begingroup$

    Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
    Thus, $g$ is convex on $big[-ln(3),inftybig)$.



    Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
    If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
    $$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
    If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
    Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't get it. What is the difference with my solution?
      $endgroup$
      – greedoid
      Dec 4 '18 at 21:45










    • $begingroup$
      I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
      $endgroup$
      – Batominovski
      Dec 4 '18 at 21:48
















    4












    4








    4





    $begingroup$

    Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
    Thus, $g$ is convex on $big[-ln(3),inftybig)$.



    Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
    If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
    $$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
    If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
    Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.






    share|cite|improve this answer











    $endgroup$



    Let $g(t):=left(dfrac{1}{1+exp(t)}right)^3$ for $tinmathbb{R}$. Then, $$g''(t)=frac{3,exp(t),big(3,exp(t)-1big)}{big(1+exp(t)big)^5}text{ for each }tinmathbb{R},.$$
    Thus, $g$ is convex on $big[-ln(3),inftybig)$.



    Let $x:=dfrac{b}{a}$, $y:=dfrac{c}{b}$, and $z:=dfrac{a}{c}$ be as the OP defines. Then, the required inequality is equivalent to
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq dfrac{3}{8},.tag{*}$$
    If $x$, $y$, or $z$ is less than $dfrac{1}{3}$, then clearly the left-hand side of (*) is greater than
    $$left(dfrac{1}{1+frac13}right)^3=frac{27}{64}>frac38,.$$
    If all $x$, $y$, and $z$ are greater than or equal to $dfrac13$, then $ln(x),ln(y),ln(z)geq -ln(3)$, so that we can use convexity of $g$ on $big[-ln(3),inftybig)$. By Jensen's Inequality,
    $$gbig(ln(x)big)+gbig(ln(y)big)+gbig(ln(z)big)geq3,gleft(frac{ln(x)+ln(y)+ln(z)}{3}right)=3,g(0)=frac{3}{8},.$$
    Hence, the equality holds if and only if $x=y=z=1$, making $a=b=c$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 4 '18 at 21:46

























    answered Dec 4 '18 at 21:42









    BatominovskiBatominovski

    1




    1












    • $begingroup$
      I don't get it. What is the difference with my solution?
      $endgroup$
      – greedoid
      Dec 4 '18 at 21:45










    • $begingroup$
      I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
      $endgroup$
      – Batominovski
      Dec 4 '18 at 21:48




















    • $begingroup$
      I don't get it. What is the difference with my solution?
      $endgroup$
      – greedoid
      Dec 4 '18 at 21:45










    • $begingroup$
      I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
      $endgroup$
      – Batominovski
      Dec 4 '18 at 21:48


















    $begingroup$
    I don't get it. What is the difference with my solution?
    $endgroup$
    – greedoid
    Dec 4 '18 at 21:45




    $begingroup$
    I don't get it. What is the difference with my solution?
    $endgroup$
    – greedoid
    Dec 4 '18 at 21:45












    $begingroup$
    I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:48






    $begingroup$
    I'm not sure what to say to that, but my idea was to bypass the inequality $frac{x+y+z}{3}geq sqrt[3]{xyz}$ that you would need to use in your attempt.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 21:48













    5












    $begingroup$

    By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
    Thus, it's enough to prove that
    $$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
    $$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
    Now, by C-S
    $$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
    Thus, it's enough to prove that
    $$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
    which is true even for all reals $a$, $b$ and $c$.



    Indeed, the last inequality is symmetric inequality by degree four,



    which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )



    it's enough to prove the last inequality for equality case of two variables and since



    it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
    $$(a-1)^2(a+3)^2geq0.$$
    Done!






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
      Thus, it's enough to prove that
      $$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
      $$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
      Now, by C-S
      $$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
      Thus, it's enough to prove that
      $$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
      which is true even for all reals $a$, $b$ and $c$.



      Indeed, the last inequality is symmetric inequality by degree four,



      which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )



      it's enough to prove the last inequality for equality case of two variables and since



      it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
      $$(a-1)^2(a+3)^2geq0.$$
      Done!






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
        Thus, it's enough to prove that
        $$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
        $$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
        Now, by C-S
        $$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
        Thus, it's enough to prove that
        $$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
        which is true even for all reals $a$, $b$ and $c$.



        Indeed, the last inequality is symmetric inequality by degree four,



        which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )



        it's enough to prove the last inequality for equality case of two variables and since



        it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
        $$(a-1)^2(a+3)^2geq0.$$
        Done!






        share|cite|improve this answer









        $endgroup$



        By Holder $$left(sum_{cyc}frac{a^3}{(a+b)^3}right)^2sum_{cyc}1geqleft(sum_{cyc}sqrt[3]{left(frac{a^3}{(a+b)^3}right)^2cdot1}right)^3=left(sum_{cyc}frac{a^2}{(a+b)^2}right)^3.$$
        Thus, it's enough to prove that
        $$frac{left(sumlimits_{cyc}frac{a^2}{(a+b)^2}right)^3}{3}geqfrac{9}{64}$$ or
        $$sumlimits_{cyc}frac{a^2}{(a+b)^2}geqfrac{3}{4}.$$
        Now, by C-S
        $$sumlimits_{cyc}frac{a^2}{(a+b)^2}=sumlimits_{cyc}frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}geqfrac{left(sumlimits_{cyc}(a^2+ab)right)^2}{sumlimits_{cyc}(a+b)^2(a+c)^2}.$$
        Thus, it's enough to prove that
        $$4left(sumlimits_{cyc}(a^2+ab)right)^2geq3sumlimits_{cyc}(a+b)^2(a+c)^2,$$
        which is true even for all reals $a$, $b$ and $c$.



        Indeed, the last inequality is symmetric inequality by degree four,



        which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )



        it's enough to prove the last inequality for equality case of two variables and since



        it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
        $$(a-1)^2(a+3)^2geq0.$$
        Done!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 21:35









        Michael RozenbergMichael Rozenberg

        101k1591193




        101k1591193























            5












            $begingroup$

            This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.



            Added:



            We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$



            The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              It seems to me there's enough here to post as an answer, no need to apologize.
              $endgroup$
              – David K
              Dec 4 '18 at 21:11
















            5












            $begingroup$

            This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.



            Added:



            We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$



            The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              It seems to me there's enough here to post as an answer, no need to apologize.
              $endgroup$
              – David K
              Dec 4 '18 at 21:11














            5












            5








            5





            $begingroup$

            This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.



            Added:



            We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$



            The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.






            share|cite|improve this answer











            $endgroup$



            This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.



            Added:



            We can rewrite the condition as $xyz=1$ and $frac{(1+x)^4}{x} = frac{(1+y)^4}{y} = frac{(1+z)^4}{z}$



            The function $g(x)=frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $infty$. This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$. It remains to solve $frac{(1+x)^4}{x} = frac{(1+1/x^2)^4}{1/x^2}$. This time, it's not hard to check $x=1$ is the only solution and we are done.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 21:44

























            answered Dec 4 '18 at 20:56









            zoidbergzoidberg

            1,065113




            1,065113








            • 2




              $begingroup$
              It seems to me there's enough here to post as an answer, no need to apologize.
              $endgroup$
              – David K
              Dec 4 '18 at 21:11














            • 2




              $begingroup$
              It seems to me there's enough here to post as an answer, no need to apologize.
              $endgroup$
              – David K
              Dec 4 '18 at 21:11








            2




            2




            $begingroup$
            It seems to me there's enough here to post as an answer, no need to apologize.
            $endgroup$
            – David K
            Dec 4 '18 at 21:11




            $begingroup$
            It seems to me there's enough here to post as an answer, no need to apologize.
            $endgroup$
            – David K
            Dec 4 '18 at 21:11


















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