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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.

Victor Loh - 3 years, 4 months ago

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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.

Sean Ty - 3 years, 4 months ago

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Yeah :D

Victor Loh - 3 years, 4 months ago

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

Sean Ty - 3 years, 4 months ago

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

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

Mathh Mathh - 3 years, 4 months ago

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