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Let us consider three sequences $(a_n)_{n\ge1}$, $(b_n)_{n\ge1}$ and $(c_n)_{n\ge1}$ having the properties:

  • $a_{n},\ b_{n},\ c_{n}\in\left(0,\ \infty\right)$
  • $a_{n}+b_{n}+c_{n}\ge\frac{a_{n}}{b_{n}}+\frac{b_{n}}{c_{n}}+\frac{c_{n}}{a_{n}}\ \forall n\ge1$
  • $\lim\limits_{n\to\infty}a_{n}b_{n}c_{n}=1 $

Prove that $$\lim_{n\to\infty}\frac{a_{n}+b_{n}+c_{n}}{a_{n}b_{n}+b_{n}c_{n}+c_{n}a_{n}}=1 $$

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I first guessed that $a_n,b_n,c_n\to1$, but now I think it might be wrong. I can only show that, when $abc=1$, we have $a/b+b/c+c/a\ge a+b+c$. It suffics to prove that $a/b+a/b+b/c\ge3a$, where AM-GM works. – Frank Science Jun 24 '12 at 16:33
What is the purpose of the word "own" in the title? – Byron Schmuland Jun 25 '12 at 16:02
@ByronSchmuland Maybe he's finding a pretty proof for his own result. – Frank Science Jun 26 '12 at 3:20
@ClaudiuMindrila As you can see, in our community, the information that your problem is original is not posted by a code word in the title. That information, and the fact that you have a proof and are looking for alternatives goes into the body of the question. – Byron Schmuland Jun 26 '12 at 18:59
@ClaudiuMindrila I should add: welcome to MSE! You have nice problems. – Byron Schmuland Jun 26 '12 at 19:01

A proof that isn't very clever but does the job: let $$f(a,b,c)=ab\left(a+b+c-\tfrac ab-\tfrac bc-\tfrac ca\right)$$ which is chosen so that $f(a_n,b_n,c_n)\ge 0$.

The idea is:

  1. Reduce to the $abc=1$ case by defining $g(a,b)=f(a,b,\frac{1}{ab})$ and proving that $f(a_n,b_n,c_n)-g(a_n,b_n)\to 0$.
  2. Prove that $g$ is nonpositive and that $g(a_n,b_n)\to 0$ implies $(a_n,b_n)\to (1,1)$.
  3. Conclude that $(a_n,b_n,c_n)\to (1,1,1)$.


  1. Since the problem is invariant under cyclic permutation of $a_n,b_n,c_n$ for any single $n$, we can assume that $c_n$ is the maximum. We get $$c_n/a_n\le a_n+b_n+c_n\le 3c_n\\ 1/a_n\le 3$$ This implies $\limsup b_n c_n\le 3$, so that $$\limsup a_n^2 b_n^3\le \limsup c_n^2 b_n^{5/2} c_n^{1/2}\le 3^{5/2}$$ So: $$a_n^2 b_n^3\left(\tfrac{1}{a_nb_nc_n}-1\right)+\tfrac{1}{a_n}(a_nb_nc_n-1)\to 0$$ and therefore $$f(a_n,b_n,c_n)-g(a_n,b_n)\to 0$$

  2. A straightforward calculation shows that $g(a,b)$ is the cubic polynomial $$-(ab)^3+(ab)^2+a^3b-a^3+a-1$$ with discriminant in $a$ $$-(b-1)^2(b+1)(23b^3+5b^2-27b+23)$$ so that, on $D=[0,+\infty)^2$, $g(a,b)$ is always non-zero when $b\ne 1$, and when $b=1$ it is non-zero only when $a=1$. Therefore because $g(a,b)$ is negative at $(0,0)$ it is everywhere negative except at $(a,b)=(1,1)$.

    Furthermore the $\sup$ of $g$ over $D$ minus any neighborhood of $(1,1)$ is negative because $D$ is closed and $g\le -1+\varepsilon$ in a neighborhood of $\infty$. As a consequence, if $g(a_n,b_n)\to 0$ then $(a_n,b_n)\to (1,1)$.

  3. Because $f(a_n,b_n,c_n)\ge 0$ and $g$ is nonpositive, $g(a_n,b_n)\to 0$, so that $(a_n,b_n)\to (1,1)$ and therefore: $$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\lim_{n\to\infty} c_n=1$$ which is actually a stronger form of the theorem.

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