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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...$x,y,z >0$ and $x^2 + y^2 + z^ = 1$, show that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
And an 'elegant way' to do that is to use spherical coordinates...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
$\chi$ $\sigma$