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Let $x,y$ and $z$ be positive numbers such that $xy+yz+zx=1$. Prove that (using Hölder's inequality) :

$$\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$$

Thanks :)

What I try:

$$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$$ So we have to prove that :

$$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $$ or

$$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow$$ $$xyz\left(x+y+z\right) \leq \frac{1}{3},$$ but I don't know if this can help me to prove the inequality.

Thanks )

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Nice inequalities! (+1) –  Chris's sis Sep 10 '12 at 20:31

1 Answer 1

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I'm using the Hölder's inequality $$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$$ So we have to prove that :

$$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $$ or

$$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow 3xyz(x+y+z) \leq \left(xy+yz+zx\right)^2=1^2$$ And this is true because

$$x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2\geq 3x^2yz+3xy^2z+3xyz^2$$ $$x^2y^2+y^2z^2+z^2x^2\geq x^2yz+xy^2z+xyz^2$$ and this is the followint inequality: $$a^2+b^2+c^2 \geq ab+bc+ca$$ for: \begin{eqnarray} a&=&xy\\ b&=&yz\\ c&=&zx. \end{eqnarray}

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