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Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e \in \Bbb R$, prove that

$\sum_{cyc} {{a+abc} \over {1+ab+abcd}} \ge {{10} \over {3}}$

First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.

I don't know what to do to solve this problem. Can somebody please give me a hint?

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closed as off-topic by Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus Jan 15 at 0:58

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  • $\begingroup$ Okay. Thank you for noticing me. $\endgroup$ – coding1101 Jan 11 at 6:53
  • $\begingroup$ What does $\sum_{cyc}$ mean? $\endgroup$ – Chickenmancer Jan 11 at 7:09
  • $\begingroup$ It means that we add it by a cycle. The problem is equivalent to ${{a+abc} \over {1+ab+abcd}}+{{b+bcd} \over {1+bc+bcde}}+{{c+cde} \over {1+cd+cdea}}+{{d+dea} \over {1+de+deab}}+{{e+eab} \over {1+ea+eabc}} \ge {10 \over 3}$. $\endgroup$ – coding1101 Jan 11 at 7:13
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Yes, C-S helps!

Indeed, let $a=\frac{y}{z},$ $b=\frac{z}{y}$, $c=\frac{t}{z}$, $d=\frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.

Thus, the condition gives $e=\frac{x}{w}$ and we obtain: $$\sum_{cyc}\frac{a+abc}{1+ab+abcd}=\sum_{cyc}\frac{\frac{y}{x}+\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{t}{z}}{1+\frac{y}{x}\cdot\frac{z}{y}+\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{t}{z}\cdot\frac{w}{t}}=$$ $$=\sum_{cyc}\frac{\frac{y}{x}+\frac{t}{x}}{1+\frac{z}{x}+\frac{w}{x}}=\sum_{cyc}\frac{y+t}{x+z+w}=-5+\sum_{cyc}\frac{x+y+z+t+w}{x+z+w}=$$ $$=-5+\frac{1}{3}\sum_{cyc}(x+z+w)\sum_{cyc}\frac{1}{x+z+w}.$$ Can you end it now?

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  • $\begingroup$ Thank you for helping me! By the way, isn't $a = \frac{y}{x}$? And I think the fourth equation is $\sum_{cyc} \frac{y+t}{x+z+w}$. $\endgroup$ – coding1101 Jan 11 at 7:41
  • $\begingroup$ @coding1101 It was typo. Thank you! I fixed. You are welcome! $\endgroup$ – Michael Rozenberg Jan 11 at 7:44

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