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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.

Thanks:)

user26857
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Iuli
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    What is the notation $\sum_{cyc}$? – Willie Wong Sep 25 '12 at 09:19
  • 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; $(x + y + z)^2 / (x^2 + y^2 + z^2) \le 3$. – Christopher A. Wong Sep 25 '12 at 09:25
  • @daniel, the summation is cyclic, so the full summation is $$\frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a}$$. – Christopher A. Wong Sep 25 '12 at 10:04

3 Answers3

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Here is a very short one.

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

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

So

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

Andreas
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    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. – dezdichado Nov 11 '21 at 17:53
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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 $$ \sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3 $$

AlbertH
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  • 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 :) – Iuli Sep 25 '12 at 10:55
  • @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) – AlbertH Sep 25 '12 at 12:01
  • @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. – AlbertH Sep 25 '12 at 12:01
  • There are a number of proofs for that inequality (with explanations) at http://artofproblemsolving.com/community/c6h183916p1010959 – Andreas Mar 24 '16 at 12:13
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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!

Hans
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Michael Rozenberg
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