Inequality--is there an elegant way to solve this?

dwsmith

Well-known member
$x,y,z >0$ and $x^2 + y^2 + z^ = 1$, show that

$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$

MarkFL

Staff member
Re: inequality--is there an elegant way to solve this?

I would try using Lagrange multipliers.

chisigma

Well-known member
$x,y,z >0$ and $x^2 + y^2 + z^ = 1$, show that

$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
A way that doesn't require high level knowledege [even if non comfortable from the point od view of computation...] is fo find the point $\displaystyle (x_{0},y_{0}, z_{0})$ of minimum of the function...

$\displaystyle f(x,y,z)= x\ y\ z + \sqrt{x^{2}\ y^{2}\ + x^{2}\ z^{2} + y^{2}\ z^{2}} - \frac{4}{3}\ \sqrt{x\ y\ z\ (x + y + z)}\ (1)$

... under the hypothesis that $\displaystyle x_{0}^{2} + y_{0}^{2}+ z_{0}^{2} = 1$ and then to verify that is $\displaystyle f(x_{0},y_{0},z_{0}) \ge 0$...

Kind regards

$\chi$ $\sigma$

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chisigma

Well-known member
A way that doesn't require high level knowledege [even if non comfortable from the point od view of computation...] is fo find the point $\displaystyle (x_{0},y_{0}, z_{0})$ of minimum of the function...

$\displaystyle f(x,y,z)= x\ y\ z + \sqrt{x^{2}\ y^{2}\ + x^{2}\ z^{2} + y^{2}\ z^{2}} - \frac{4}{3}\ \sqrt{x\ y\ z\ (x + y + z)}\ (1)$

... under the hypothesis that $\displaystyle x_{0}^{2} + y_{0}^{2}+ z_{0}^{2} = 1$ and then to verify that is $\displaystyle f(x_{0},y_{0},z_{0}) \ge 0$...

Kind regards

$\chi$ $\sigma$
And an 'elegant way' to do that is to use spherical coordinates...

$\displaystyle x= r\ \sin \theta\ \cos \phi$

$\displaystyle y = r\ \sin \theta\ \sin \phi$

$z=r\ \cos \theta\ (1)$

... then evaluate the absolute minimum $\displaystyle (\theta_{0}, \phi_{0})$ of $\displaystyle f(1,\theta,\phi)$ and finally verify that $\displaystyle f(1,\theta_{0},\phi_{0}) \ge 0$...

Kind regards

$\chi$ $\sigma$

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dwsmith

Well-known member
And an 'elegant way' to do that is to use spherical coordinates...

$\displaystyle x= r\ \sin \theta\ \cos \phi$

$\displaystyle y = r\ \sin \theta\ \sin \phi$

$z=r\ \cos \theta\ (1)$

... then evaluate the absolute minimum $\displaystyle (\theta_{0}, \phi_{0})$ of $\displaystyle f(1,\theta,\phi)$ and finally verify that $\displaystyle f(1,\theta_{0},\phi_{0}) \ge 0$...

Kind regards

$\chi$ $\sigma$

When we re-write $f$, we get

\begin{align}
f(1,\theta,\phi) &= \sqrt{\sin ^2(\theta ) \left(\sin ^2(\theta ) \sin ^2(\phi ) \cos ^2(\phi )+\cos ^2(\theta )\right)}+\sin ^2(\theta ) \cos (\theta ) \sin (\phi ) \cos (\phi )\\
&-\frac{4}{3} \sqrt{\sin ^2(\theta ) \cos (\theta ) \sin (\phi ) \cos (\phi ) (\sin (\theta ) (\sin (\phi )+\cos (\phi ))+\cos (\theta ))}
\end{align}

Are there some trig identities I need to be utilizing now?