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Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:

$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$

Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true

and for even $n$ and positives $a_i$ the following inequality is true.

$$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$

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  • $\begingroup$ hello Michael nice to meet you here i think BW! $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '14 at 18:02
  • $\begingroup$ Hello, Dr. Sonnhard Graubner! I think BW is not useable here. $\endgroup$ – Michael Rozenberg Nov 11 '14 at 20:31
  • $\begingroup$ The right side is never more than $n/2$. Can the left side be less than $n/2$? $\endgroup$ – Paul Jan 23 '16 at 14:40
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    $\begingroup$ @Paul: For $(a, b, c, d, e) = (1, k, k^2, k^3, k^4)$ and $k \to \infty$, the lhs comes arbitrarity close to one. $\endgroup$ – Martin R Jan 23 '16 at 22:06
  • $\begingroup$ @Michael: Can you share your "easy proof" for $n=3$ and $n$ even? $\endgroup$ – Martin R Jan 24 '16 at 17:55
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Here is a full proof.

Let us start the discussion for general $n$. Denote $S = \sum_{i=1}^n a_i$.

Since by AM-GM, $S \geq n \sqrt[n]{a_1a_2...a_n}$, we have

$$1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} \geq \frac{S}{S - (n-2)\sqrt[n]{a_1a_2...a_n}}$$

Hence a tighter claim is (simultaneously defining $L$ and $R$):

$$L = \sum_{cyc}\frac{a_i}{a_i+a_{i+1}}\geq 1+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{2S} = R$$ and it suffices to show that one.

We write $2 L \geq 2 R$ or $L \geq 2 R- L$ and add on both sides a term $$\sum_{cyc}\frac{a_{i+1}}{a_i+a_{i+1}}$$ which leaves us to show

$$n = \sum_{cyc}\frac{a_i + a_{i+1}}{a_i+a_{i+1}}\geq 2+\frac{n(n-2)\sqrt[n]{a_1a_2...a_n}}{S} + \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}}$$ or, in our final equivalent reformulation of the L-R claim above, $$ \sum_{cyc}\frac{-a_i + a_{i+1}}{a_i+a_{i+1}} \leq (n - 2) (1- \frac{n \sqrt[n]{a_1a_2...a_n}}{S} )$$

For general odd $n$ see the remark at the bottom. Here the task is to show $n=5$.

Before doing so, we will first prove the following Lemma (required below), which is the above L-R-inequality for 3 variables (which is tighter than the original formulation, hence we cannot apply the proof for $n=3$ given above by Michael Rozenberg for the original formulation):

$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c} \leq (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} )$$

This Lemma is, from the above discussion, just a re-formulation of the claim in $L$ and $R$ above, for 3 variables, i.e.

$$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$

By homogeneity, we can demand $abc=1$ and prove, under that restriction, $$ \frac{a}{b+a} + \frac{b}{c+b} + \frac{c}{a+c} \geq 1+\frac{3}{2(a+b+c)}$$

This reformulates into $$ \frac{a\; c}{a +b} + \frac{b\; a}{b +c} + \frac{c\; b}{c +a} \geq \frac{3}{2}$$ or equivalently, due to $abc=1$, $$ \frac{1}{b(a +b)} + \frac{1}{c(b +c)} + \frac{1}{a(c +a)} \geq \frac{3}{2}$$

which is known (2008 International Zhautykov Olympiad), for some proofs see here: http://artofproblemsolving.com/community/c6h183916p1010959 Hence the Lemma holds.

For $n=5$, we rewrite the LHS of our above final reformulation by adding and subtracting terms:

$$ \frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e} = \\ (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + (\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e}) $$ This also holds for any cyclic shift in (abcde), so we can write

$$ 5 (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{d-c}{d+c} + \frac{e-d}{e+d} + \frac{a-e}{a+e}) = \\ \sum_{cyc (abcde)} (\frac{b-a}{b+a} + \frac{c-b}{c+b} + \frac{a-c}{a+c}) + \sum_{cyc (abcde)}(\frac{c-a}{c+a}+\frac{d-c}{d+c} + \frac{a-d}{a+d}) + \sum_{cyc (abcde)} (\frac{d-a}{d+a}+ \frac{e-d}{e+d} + \frac{a-e}{a+e}) $$

Using our Lemma, it suffices to show (with $S = a +b+c+d+e$)

$$ \sum_{cyc (abcde)} (1- \frac{3 \sqrt[3]{a\, b \, c}}{a + b+ c} ) + \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, c \, d}}{a + c+ d} ) + \sum_{cyc (abcde)}(1- \frac{3 \sqrt[3]{a\, d \, e}}{a + d+ e} ) \leq 15 (1- \frac{5 \sqrt[5]{a b c d e }}{S} ) $$ which is $$ \sum_{cyc (abcde)} (\frac{\sqrt[3]{a\, b \, c}}{a + b+ c} + \frac{\sqrt[3]{a\, c \, d}}{a + c+ d} + \frac{\sqrt[3]{a\, d \, e}}{a + d+ e} ) \geq 25 \frac{\sqrt[5]{a b c d e }}{S} $$ Using Cauchy-Schwarz leaves us with showing $$ \frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2}{\sum_{cyc (abcde)}(a + b+ c)} + \frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2}{\sum_{cyc (abcde)}(a + c+ d)} + \frac {(\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2}{\sum_{cyc (abcde)}(a + d+ e)} \geq 25 \frac{\sqrt[5]{a b c d e }}{S} $$ The denominators all equal $3S$, so this becomes $$ (\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2 + (\sum_{cyc (abcde)} \sqrt[6]{a\, c \, d})^2 + (\sum_{cyc (abcde)} \sqrt[6]{a\, d \, e})^2 \geq 75 \sqrt[5]{a b c d e } $$ Using AM-GM gives for the first term

$$ (\sum_{cyc (abcde)} \sqrt[6]{a\, b \, c})^2 \geq ( 5 (\prod_{cyc (abcde)} \sqrt[6]{a\, b \, c} )^{1/5})^2 = 25 (\prod_{cyc (abcde)} ({a\, b \, c} ) )^{1/15} = 25 \sqrt[5]{a b c d e } $$

By the same procedure, the second and the third term on the LHS are likewise greater or equal than $25 \sqrt[5]{a b c d e }$. This concludes the proof.

Remarks:

  1. the tighter $L-R$-claim used here is - for general $n$ - asked for in the problem given at Cyclic Inequality in n (at least 4) variables

  2. For general odd $n$, the above reformulation can be used again. For odd $n>5$, take the method of adding and subtracting terms to form smaller sub-sums which are cyclically closed in a smaller number of variables, and apply previous results for smaller $n$ recursively.

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  • $\begingroup$ Beautiful proof. Thank you! $\endgroup$ – Michael Rozenberg Apr 29 '16 at 12:14
  • $\begingroup$ @Michael Rozenberg. Thank you for the bounty! In the original question you wrote "for even n and positives $a_i$ the following inequality is true." Could you make that solution available? $\endgroup$ – Andreas Apr 29 '16 at 12:47
  • $\begingroup$ The proof for any $n$ (for the tighter L-R-claim stated above) is given here: math.stackexchange.com/questions/1711515 $\endgroup$ – Andreas Apr 29 '16 at 18:33
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A proof for $n=3$.

We'll prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$ for all positives $a$, $b$ and $c$.

Indeed, let $ab+ac+bc\geq(a+b+c)\sqrt[3]{abc}$.

Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}=\frac{1}{1-\frac{ab+ac+bc}{(a+b+c)^2}}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.

Let $ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$.

Hence, by C-S $\sum\limits_{cyc}\frac{a}{a+b}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2c^2+a^2bc)}=\frac{1}{1-\frac{abc(a+b+c)}{(ab+ac+bc)^2}}\geq\frac{1}{1-\frac{\sqrt[3]{abc}}{a+b+c}}=\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$.

Done!

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A proof for even $n$.

Let $a_i>0$, $a_{n+1}=a_1$ and $n$ is an even natural number. Prove that: $$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$ Proof.

By C-S and AM-GM $\sum\limits_{i=1}^n\frac{a_i}{a_i+a_{i+1}}=\sum\limits_{k=1}^{\frac{n}{2}}\frac{a_{2k-1}}{a_{2k-1}+a_{2k}}+\sum\limits_{k=1}^{\frac{n}{2}}\frac{a_{2k}}{a_{2k}+a_{2k+1}}\geq\frac{\left(\sum\limits_{k=1}^{\frac{n}{2}}\sqrt{a_{2k-1}}\right)^2}{a_1+a_2+...+a_n}+\frac{\left(\sum\limits_{k=1}^{\frac{n}{2}}\sqrt{a_{2k}}\right)^2}{a_1+a_2+...+a_n}\geq$

$\geq\frac{a_1+a_2+...+a_n+\frac{n^2-2n}{2}\sqrt[n]{a_1a_2...a_n}}{a_1+a_2+...+a_n}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$.

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  • $\begingroup$ Thanks for the proof for even $n$ - very short and concise! $\endgroup$ – Andreas Apr 29 '16 at 21:59
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I'm going to prove a different but similar inequality.

Let $x_1,x_2,\ldots,x_n>0$, $x_{n+1}=x_1$, $x_{n+2}=x_2$, $n$ be a positive integer such that either

$a)\ $ $n\le 12$; or

$b)\ $ $13\le n\le 23$, $n$ is odd.

Then this inequality is true:

$$\sum_{i=1}^n\frac{x_i}{x_{i+1}+x_{i+2}}\ge \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i - (n-2)\sqrt[n]{\prod_{i=1}^n x_i}}$$

Proof: by AM-GM:

$$\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i - (n-2)\sqrt[n]{\prod_{i=1}^n x_i}}\le $$

$$\le \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i - (n-2)\frac{\sum_{i=1}^n x_i}{n}}=$$

$$=\frac{\sum_{i=1}^n x_i}{\frac{2}{n}\sum_{i=1}^n x_i}=\frac{n}{2}$$

By Shapiro inequality: $$\frac{n}{2}\le \sum_{i=1}^n\frac{x_i}{x_{i+1}+x_{i+2}}$$

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