An inequality proposed at Zhautykov Olympiad 2008.

Let be $a,b,c >0$ with $abc=1$. Prove that: $$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$

Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.

Our inequality becomes: $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$ Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$

Now applying Cauchy-Schwarz we obtain the desired result.

What I wrote can be found on this link: mateforum. But now, I don't know how to apply Cauchy-Schwarz.


  • 2
    $\begingroup$ What is the notation $\sum_{cyc}$? $\endgroup$ Commented Sep 25, 2012 at 9:19
  • $\begingroup$ Tell me if this is where you get stuck: If $S$ is the last sum you have above, then using Cauchy-Schwarz you can obtain the inequality $$\tfrac{1}{2}S(4x^2 + 4y^2 + 4z^2) \ge (x + y + z)^2.$$ What's disturbing about this is that you then get $S \ge \tfrac{1}{2} (x + y + z)^2/(x^2 + y^2 + z^2)$. It would be natural then to try to prove that $(x + y + z)^2 / (x^2 + y^2 + z^2) \ge 3$, but instead the reverse is true; $(x + y + z)^2 / (x^2 + y^2 + z^2) \le 3$. $\endgroup$ Commented Sep 25, 2012 at 9:25
  • $\begingroup$ @daniel, the summation is cyclic, so the full summation is $$\frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a}$$. $\endgroup$ Commented Sep 25, 2012 at 10:04

3 Answers 3


Here is a very short one.

Suppose w.l.o.g. $a \ge b \ge c$. Then by the rearrangement inequality,

$$ S = \frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a} \ge \frac{1}{(a+b)c} +\frac{1}{(b+c)a} + \frac{1}{(a+c)b} = T $$


$$ 2 S \ge S + T = \frac{b+c}{(a+b)bc} +\frac{c+a}{(b+c)ca} + \frac{a+b}{(a+c)ab} \geq 3 \Big( \frac{1}{abc} \Big)^{2/3} = 3 $$ which proves it, where in the last step AM-GM was used.

  • $\begingroup$ Technically, one cannot assume $a\geq b\geq c$ as the expression is not symmetric. I would say no matter the ordering of $a,b,c$, $S\geq T$ due to rearrangement. $\endgroup$
    – dezdichado
    Commented Nov 11, 2021 at 17:53

Since $\mathrm{LHS}$ of last inequality is homogeneous we can assume $x^2 + y^2 + z^2 = 1$. Then it becomes $$ \mathrm{LHS} = 2\sum_{cyc} \frac {x^2} {1 + z^2} =:2I $$ Now using Cauchy-Schwarz inequality we get $$ 1 = (x^2 + y^2 + z^2)^2 = \left(\sum_{cyc} x\sqrt{1 + z^2} \cdot \frac x {\sqrt{1 + z^2}}\right)^2 \leq\\ \left(\sum_{cyc} x^2(1 + z^2)\right) \cdot \left( \sum_{cyc} \frac {x^2}{1 + z^2} \right) = I \cdot \sum_{cyc} x^2(1 + z^2) $$ To finish, let's note that CS inequality implies $$ x^2\cdot z^2 + y^2 \cdot x^2 + z^2 \cdot y^2 \leq x^4 + y^4 + z^4 $$ and therefore $$ \sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3 $$

  • $\begingroup$ I have a question :) How do you know to apply the relation of homogenity and how do you apply it ? Why $x^2+y^2+z^2=1? $ Thanks :) $\endgroup$
    – Iuli
    Commented Sep 25, 2012 at 10:55
  • $\begingroup$ @Iuli Since $\mathrm{LHS}(x, y, z) = \mathrm{LHS}(kx, ky, kz)$, if the inequality holds for $(x, y, z)$, it holds for each $(kx, ky, kz)$ with $k > 0$. So, for example, if you want to prove the inequality for $(x, y, z)$, it is sufficient you prove it for $(kx, ky, kz)$ with $k = 1/\sqrt{x^2 + y^2 + z^2}$ and that is equivalent to assume the condition $x^2 + y^2 + z^2 = 1$. Of course, you are free to choose other values for $k$ and therefore other conditions on $(x, y, z)$. $k = 1/\sqrt[3]{abc}$, for example, is equivalent to the condition $abc=1$. (continue) $\endgroup$
    – AlbertH
    Commented Sep 25, 2012 at 12:01
  • $\begingroup$ @Iuli I used the condition $x^2 + y^2 + z^2 = 1$ because I felt it would have simplified the structure of the inequality and (overall) reduced the amount of symbols to write. However, the line of the proof remains the same also without the above assumption. $\endgroup$
    – AlbertH
    Commented Sep 25, 2012 at 12:01
  • $\begingroup$ There are a number of proofs for that inequality (with explanations) at artofproblemsolving.com/community/c6h183916p1010959 $\endgroup$
    – Andreas
    Commented Mar 24, 2016 at 12:13

Let $a=\frac{x}{y}$ and $b=\frac{y}{z}$, where $x$, $y$ and $z$ be positives.

Hence, since $abc=1$, we get $c=\frac{z}{x}$ and by C-S we obtain: $$\sum_{cyc}\frac{1}{(a+b)b}=\sum_{cyc}\frac{z^2}{xz+y^2}=\sum_{cyc}\frac{z^4}{xz^3+z^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3y)}.$$ Thus, it remains to prove that $$2(x^2+y^2+z^2)^2\geq3\sum\limits_{cyc}(x^2y^2+x^3y)$$ or $$\sum_{cyc}(2x^4-3x^3y+x^2y^2)\geq0$$ or $$\sum_{cyc}(x-y)x^2(2x-y)\geq0$$ or $$\sum_{cyc}\left((x-y)x^2(2x-y)-\frac{x^4-y^4}{4}\right)\geq0$$ or $$\sum_{cyc}(x-y)^2(7x^2+2xy+y^2)\geq0.$$ Done!

  • $\begingroup$ Nice trick on the penultimate line. $\endgroup$
    – Hans
    Commented Sep 3, 2017 at 2:18
  • 1
    $\begingroup$ As an aside, expanding the "remains to prove" results in $ \sum 2x^2 + \sum x^2y^2 \geq 3 \sum x^3 y$, which is true by AM-GM. $\endgroup$
    – Calvin Lin
    Commented Sep 30, 2021 at 16:25
  • $\begingroup$ @Calvin Lin Yes, I agree. Thank you for your interest! $\endgroup$ Commented Sep 30, 2021 at 16:38

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