\(abc=1\)

(IMO 1995)\textbf{(IMO 1995)}

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

1a3(b+c)+1b3(c+a)+1c3(a+b)≥32.\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.

#Algebra

Note by Victor Loh
6 years, 10 months ago

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Comments

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

∑1a3(b+c)=∑11x3(1y+1z)=∑x2y+z.\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

∑x2y+z≥(x+y+z)22(x+y+z)=x+y+z2≥3xyz32=32,\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 - 6 years, 10 months ago

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

By Cauchy-Schwarz in Engel Form, we have

∑cyc1a3(b+c)≥(1a+1b+1c)22(ab+bc+ca)\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 - 6 years, 10 months ago

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Yeah :D

Victor Loh - 6 years, 10 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 - 6 years, 10 months ago

∑cyc1a3(b+c)=∑cycb2c2a(b+c)\sum_{\text{cyc}} \frac{1}{a^3(b+c)}=\sum_{\text{cyc}} \frac{b^2c^2}{a(b+c)}

≥Cauchy-Schwarz(ab+bc+ca)22(ab+bc+ca)=ab+bc+ca2≥AM-GM32\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 - 6 years, 10 months ago
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