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let $a,b,c,d,e$ are positive real numbers,show that $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+e}+\dfrac{d-e}{e+a}+\dfrac{e-a}{a+b}\ge 0$$

My try:

I have solved follow Four-variable inequality:

let $a,b,c,d$ are positive real numbers,show that $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge 0$$ poof:since \begin{align*} &\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}=\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}-4\\ &=(a+c)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)+(b+d)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)-4 \end{align*} By Cauchy-Schwarz inequality we have $$\dfrac{1}{b+c}+\dfrac{1}{d+a}\ge\dfrac{4}{(b+c)+(d+a)},\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{(c+d)+(a+b)}$$ so we get $$\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge\dfrac{4(a+c)}{(b+c)+(d+a)}+\dfrac{4(b+d)}{(c+d)+(a+b)}-4=0$$ Equality holds for $a=c$ and $b=d$

By done!

But for Five-variable inequality,I can't prove it.Thank you

math110
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  • Why do you believe it is true? Where does this come from? – Igor Rivin Dec 30 '13 at 03:46
  • Just a side note: instead of using Cauchy-Schwarz, you can also use (the IMO less complicated) Arithmetic-Harmonic Mean inequality. Have you solved the inequality for $n=3$? ?(I suspect the proof will be similar to the $n=5$ case.) – Ragnar Dec 30 '13 at 03:55
  • I suspect multiplying with all denominators and using Muirhead will work, but that is not a solution that can be done by hand. Otherwise, using reordering may work. – Ragnar Dec 30 '13 at 04:05
  • A proof for $n=3$ is possible with convexity, however that proof does not work analogously for $n=5$. – Andreas Dec 11 '18 at 08:41
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    Yang Lu and Xia Bican, in their book "Automated Inequality Proving And Discovering" (World Scientific, 2016, ISBN 9814759139, 9789814759137) , have shown (see page 291, examples 10.19 and 10.20) by Successive Difference Substitution that the inequality holds, indeed it does not hold in 6 variables and it holds again in 7 variables. See here: https://books.google.de/books?id=MGC2DAAAQBAJ&pg=PA291&lpg=PA291&dq=conjecture+cirtoaje&source=bl&ots=VgwaAhwvRf&sig=dK85ICa8kt37dDTWh05vVmxyfkg&hl=de&sa=X&ved=2ahUKEwiy9q7A_5ffAhWiWhUIHamZAmw4FBDoATAAegQICRAB#v=snippet&q=cirtoaje&f=false – Andreas Dec 16 '18 at 19:13

2 Answers2

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For $n$ variables $x_i > 0$, (with the convention $x_{n+k}=x_k$) we need to show that $$\sum_i \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} \geqslant 0 \iff \sum_i \left( \dfrac{x_i - x_{i+1}}{x_{i+1} + x_{i+2}} +\frac12 \right) \geqslant \frac{n}2 \iff \sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant n $$

By Cauchy-Schwarz we have $$\sum_i \dfrac{2x_i - x_{i+1}+x_{i+2}}{x_{i+1} + x_{i+2}} \geqslant \dfrac{ \left(\sum_i (2x_i - x_{i+1}+x_{i+2}) \right)^2}{\sum_i \left( (x_{i+1} + x_{i+2}) (2x_i - x_{i+1}+x_{i+2} ) \right)} = \dfrac{ 4\left(\sum_i x_i \right)^2}{2\sum_i (x_i x_{i+1} + x_i x_{i+2})} $$

So it is sufficient to show that $$ \left(\sum_i x_i \right)^2 \geqslant \dfrac{n}2 \sum_i x_i (x_{i+1}+x_{i+2}) \tag{*}$$
While the general case eludes me, this helps to answer your question and the case $n=6$.

For $n=5$ we have the SOS form: $$\left(\sum_{i=1}^5 x_i \right)^2 - \dfrac{5}2 \sum_{i=1}^5 x_i (x_{i+1}+x_{i+2}) = \frac14 \sum_{i=1}^5 (x_i - x_{i+1})^2 + \frac14 \sum_{i=1}^5 (x_i - x_{i+2})^2 \geqslant 0$$ For equality we will need all $x_i$ to be the same.

For $n=6$, we can use $a=x_1 + x_4, b = x_2 + x_5, c = x_3+x_6$ to get: $$\left(\sum_{i=1}^6 x_i \right)^2 - \dfrac{6}2 \sum_{i=1}^6 x_i (x_{i+1}+x_{i+2}) = (a+b+c)^2 - 3(ab+bc+ca) \geqslant 0$$ In this case for equality we need all $x_i$ for odd $i$ to be the same, and similarly all the $x_i$ for even $i$ must also be the same.

Added based on Dongryul Kim's comment - this application of Cauchy Schwarz requires an additional condition like $x_i \in [\frac1{\sqrt3}, \sqrt3]$ so that the numerators in the fraction on LHS are non-negative.

Macavity
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  • Oh,It's very nice !Thank you – math110 Dec 30 '13 at 12:41
  • Is the inequality $$\left(\sum_{i=1}^n x_i\right)^2\geq \frac{n}{2}\sum_{i=1}^nx_i(x_{i+1}+x_{i+2})$$ true for any positive numbers $x_1,x_2,\cdots,x_n$? According to Macavity, it is not always true. But I can not find the counter-case. Can anyone give a such case? – nuage Jan 03 '14 at 12:39
  • @azhi Try $(1, 0, 1, 0, 1, 0)$. Note that being strictly positive doesn't help, as we could replace $0$ with an arbitrarily small real number. – Macavity Jan 03 '14 at 13:03
  • There exist positive numbers $x_1,\cdots,x_n,$ such that $$\left(\sum_{i=1}^n x_i\right)^2\not\geq \frac{n}{2}\sum_{i=1}^n x_i(x_{i+1}+x_{i+2}).$$ Here is a counterexample: $n=8, (x_1,x_2,\cdots,x_8)=(1,2,3,4,4,3,2,1).$ – nuage Jan 04 '14 at 09:14
  • I gave a counter example earlier. That inequality does not hold for any $n>6$. – Macavity Jan 04 '14 at 09:56
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    @Macavity Your solution is wrong because $2x_i-x_{i+1}+x_{i+2}$ can be negative. – Michael Rozenberg Dec 15 '16 at 00:41
  • @MichaelRozenberg Isnt that exactly what was discussed in the other post, comments and addendum above? Whats new? – Macavity Dec 15 '16 at 06:14
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    @Macavity The new is why you post a wrong solution? I think it will be better if you'll delete your wrong solution. Thank you! – Michael Rozenberg Dec 15 '16 at 12:23
  • @MichaelRozenberg often on this site partial solutions and attempts are preserved so as to inform others. Please note it's is clear the solution is not complete, and if you have a solution please provide, the OP will accept it. – Macavity Dec 15 '16 at 12:26
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    @Macavity I understand you (I don't agree with you with a term "solution" because I think that not complete solution is not solution, but I understand you). What that I don't understand, why not complete solution gets many positive votes and my (and not only my) full solutions do not get any vote? Why it happens? – Michael Rozenberg Dec 15 '16 at 12:37
  • If you're worry is I got votes, I shall make my partial answer a community one, so that I don't benefit. If your worry is that you didn't get enough credit, all I can do is to vote up any i see that merits an up vote. I don't always agree with your methods e.g. $uvw$ because they are obscure and I have yet to see proper proofs. Other ones glad to upvote. – Macavity Dec 15 '16 at 12:40
  • Ok it seems I can't change to community this one perhaps it's too old. – Macavity Dec 15 '16 at 12:42
  • @MichaelRozenberg On the other hand I just noticed you get a lot of upvotes just by posting unsolved problems, so keep at it, I am sure you will get credit there if not for answers. – Macavity Dec 15 '16 at 12:45
  • @Macavity Thank you! By the way, the $uvw$ is not obscure method. I am ready to explain all steps. – Michael Rozenberg Dec 15 '16 at 20:13
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    I found this interesting paper http://www.turgor.ru/lktg/2010/5/5-1en.pdf –  Aug 10 '17 at 14:34
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I would have written this in a comment, but I don't have enough 'reputation' to do so...

@Macavity I'd just like to point out that $$\sum \frac{a_i}{b_i} \ge \frac{\big(\sum a_i \big)^2}{\sum a_i b_i}$$ holds when $a_i b_i$ is positive for all $i$. (To use Cauchy-Schwarz, you need to square-root and then square those terms) A counterexample for this might be $a_1=-1, a_2=1, b_1=2$, and $b_2=4$.

Also, this is a conjecture proposed by Vasile Cirtoaje many years ago in Old and New Inequalities (page 60). Inequalities with more variables were discussed here

Dongryul Kim
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