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# $$abc=1$$

$$\textbf{(IMO 1995)}$$

Let $$a, b, c$$ be positive real numbers such that $$abc=1$$. Prove that

$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$

Note by Victor Loh
3 years, 4 months ago

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Let $$x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$$. Then by the given condition we obtain $$xyz=1$$. Note that

$\sum \frac{1}{a^3(b+c)}=\sum \frac{1}{\frac{1}{x^3}(\frac{1}{y}+\frac{1}{z})}=\sum \frac{x^2}{y+z}.$

Now by Cauchy-Schwarz inequality

$\sum \frac{x^2}{y+z} \geq \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2} \geq \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2},$

where the last inequality follows from AM-GM.

- 3 years, 4 months ago

You may use Cauchy-Schwarz in Engel Form as well.

By Cauchy-Schwarz in Engel Form, we have

$\sum_{cyc} \dfrac{1}{a^{3}(b+c)} \geq \dfrac{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^{2}}{2(ab+bc+ca)}$

And you can continue from here.

- 3 years, 4 months ago

Yeah :D

- 3 years, 4 months ago

Did you see this from a book? :D (if I won't reply, I'm in school :) )

- 3 years, 4 months ago

$\sum_{\text{cyc}} \frac{1}{a^3(b+c)}=\sum_{\text{cyc}} \frac{b^2c^2}{a(b+c)}$

$\stackrel{\textbf{Cauchy-Schwarz}}\ge \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\stackrel{\textbf{AM-GM}}\ge \frac{3}{2}$

- 3 years, 4 months ago