Inequality

**PQR** · November 6th 2009, 01:12 PM

For any $a, b, c > 0$ with $abc = 1$ , prove that

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

**bigwave** · November 6th 2009, 01:38 PM

with abc = 1
then $a=1\ b=1 \ and \ c=1$

therefore

$ \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} \geq \frac{3}{2} $

**Krizalid** · November 6th 2009, 01:59 PM

you don't have a proof, that's a particular case you just took.

**PQR** · November 10th 2009, 06:05 AM

Originally Posted by bigwave

with abc = 1
then $a=1\ b=1 \ and \ c=1$

therefore

$ \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} + \frac{1}{1^3(1+1)} \geq \frac{3}{2} $

If it only was that easy!

**Krizalid** · January 4th 2010, 04:04 PM

$a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ac)=2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right),$ and $ab+bc+ac\ge 3\sqrt[3]{a^{2}b^{2}c^{2}}=3,$ now by Cauchy - Schwarz we have $ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1} {c^{3}(a+b)} \right)2(ab+bc+ac)\ge \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^{2},$ finally $\left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1} {c^{3}(a+b)} \right)\ge \frac{ab+bc+ac}{2}\ge \frac{3}{2}.\quad\blacksquare$

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