Back To Basics II

Since the constraints are squares of numbers, and we aren't subtracting or otherwise doing anything to the expression where a negative sign would matter, we can assume that $x,y,z$ are all positive.

Rewrite our constraint as

$x^{2}+\frac{1}{2}y^{2}+\frac{1}{2}y^{2}+z^{2}=1$

By AM-GM Inequality , we get that

$\frac{x^{2}+\frac{1}{2}y^{2}+\frac{1}{2}y^{2}+z^{2}}{4} \geq \sqrt[\Large{4}]{\frac{1}{4}x^{2}y^{4}z^{2}}$

$\frac{\sqrt{2}}{4} \geq \sqrt{xy^{2}z}$

$xy^{2}z \leq \frac{1}{8}$

Therefore

$3xy^{2}z \leq \frac{3}{8}$ .

And we know that equality can occur because by AM-GM once again, we get that

$x^{2}=\frac{1}{2}y^{2}=z^{2}$

And by our constraint we get

$4x^{2}=1 \\ 2y^{2}=1 \\ 4z^{2}=1$

All of which have solutions over the real numbers, therefore equality must be obtained.

And our answer is $3+8=\boxed{11}$

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For the generalization:

Rewrite our constraint as

$\sum _{ i=1 }^{ m }{ \frac { { x }^{ 2 } }{ m } } +\sum _{ j=1 }^{ n }{ \frac { { y }^{ 2 } }{ n } } +\sum _{ k=1 }^{ p }{ \frac { { z }^{ 2 } }{ p } } =1$

And so, by AM-GM, we get

$\frac{\sum _{ i=1 }^{ m }{ \frac { { x }^{ 2 } }{ m } } +\sum _{ j=1 }^{ n }{ \frac { { y }^{ 2 } }{ n } } +\sum _{ k=1 }^{ p }{ \frac { { z }^{ 2 } }{ p } }}{m+n+p} \geq \sqrt[\large{m+n+p}]{\frac{x^{2m}y^{2n}z^{2p}}{m^{m}n^{n}p^{p}}}$

$\frac{1}{m+n+p} \geq \sqrt[\large{m+n+p}]{\frac{x^{2m}y^{2n}z^{2p}}{m^{m}n^{n}p^{p}}}$

$\frac{m^{m}n^{n}p^{p}}{(m+n+p)^{m+n+p}} \geq x^{2m}y^{2n}z^{2p}$

And finally...

$x^{m}y^{n}z^{p} \leq \sqrt{\frac{m^{m}n^{n}p^{p}}{(m+n+p)^{m+n+p}}}$