3 variables, 2 equations, 1 answer

Method 1

$Using\quad power-mean\quad inequality,\\ \\ \cfrac { x+y }{ 2 } \le \sqrt { \cfrac { { x }^{ 2 }+{ y }^{ 2 } }{ 2 } } \\ \\ \Rightarrow { \left( \cfrac { x+y }{ 2 } \right) }^{ 2 }\le \cfrac { { x }^{ 2 }+{ y }^{ 2 } }{ 2 } \\ \\ \Rightarrow \cfrac { { z }^{ 4 } }{ 4 } \le \cfrac { { z }^{ 3 } }{ 2 } \\ \\ \Rightarrow z\le 2\quad (as\quad z\quad is\quad +ve\quad so\quad cancel\quad { z }^{ 3 })\\ \\ so\quad possible\quad +ve\quad integral\quad values\quad of\quad z\quad are\quad 1\& 2\\ \\ by\quad hit\quad \& \quad trial\quad only\quad one\quad solution\quad (x,y,z)=(2,2,2)\\ \\ \Rightarrow Ans=2+2+2=6$

Method 2

$Let(by\quad { eq }^{ n }\quad 2)\quad x={ z }^{ \cfrac { 3 }{ 2 } }\cos { \theta ,y= } { z }^{ \cfrac { 3 }{ 2 } }\sin { \theta } \\ \\ By\quad { eq }^{ n }\quad 1,\\ \\ { z }^{ \cfrac { 3 }{ 2 } }\left( \cos { \theta +\sin { \theta } } \right) ={ z }^{ 2 }\\ \\ \Rightarrow { z }^{ \cfrac { 1 }{ 2 } }=\cos { \theta +\sin { \theta } } \\ \\ \Rightarrow z=1+\sin { 2\theta } \\ \\ so\quad possible\quad +ve\quad integral\quad values\quad of\quad z\quad are\quad 1\& 2\\ \\ by\quad hit\quad \& \quad trial\quad only\quad one\quad solution\quad (x,y,z)=(2,2,2)\\ \\ \Rightarrow Ans=2+2+2=6$

This problem seems to be incorrect. There should be solutions $(x,y)$ for any $z$ with $1 < z \le 2$ . Namely, let $\begin{aligned} x &= \frac{z^2}2 \left( 1 + \sqrt{\frac2{z}-1} \right) \\ y &= \frac{z^2}2 \left( 1 - \sqrt{\frac2{z}-1} \right) \\ \end{aligned}$ For instance, if $z = 8/5$ we get $(x,y) = (48/5,16/5)$ .

Perhaps we want to restrict to positive integers? In that case, we get $x = y = z = 2$ as the only solutions.