A inequality proposed at Zhautykov Olympiad 2008

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:)

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 $$ \sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3 $$

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.

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!