Solving $\;x+y+z =8 ; \;\;\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10 $ Solve the problem 
\begin{cases}x+y+z =8 \\ \\
\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10 
\end{cases}
with $(x,y,z) \in \mathbb R^3$
I have already solved it, but I'd like to see others creative solutions and before all, share this funny problem with the community. 
 A: Think Length of path on $\mathbb{R}^2$ $(0,0)-(x,1)-(x+y,3)-(x+y+z,6)$.
Minimum length of path is 10. When those points are on one line.
$x=8/6,y=16/6,z=24/6$.
A: Since $\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}\ge \sqrt{(x+y+z)^2+(1+2+3)^2}= 10$ by the Minkowski inequality, now check when the equal sign hold.
A: (This answer uses the ideas inherent in the answers of Maddy and Ma Ming.)
Consider the three vectors $${\bf a}_1:=(x,1),\quad {\bf a}_2:=(y,2), \quad {\bf a}_3:=(z,3)$$
in ${\mathbb R}^2$. By the triangle inequality
$$|{\bf a}_1|+|{\bf a}_2|+|{\bf a}_3|>|{\bf a}_1+{\bf a}_2+{\bf a}_3|\ ,\tag{1}$$
or
$$\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}>\sqrt{(x+y+z)^2+(1+2+3)^2}\ ,$$
unless the  ${\bf a}_i$ are positive scalar multiples of each other, in which case we have equality sign in $(1)$.
Now we are given the condition $$x+y+z=8\ .\tag{2}$$ It follows that for any triple $({\bf a}_i)_{1\leq i\leq3}$ satisfying $(2)$ we necessarily have
$$\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}>\sqrt{100}=10\ ,$$
apart from potential ${\bf a}_i$ which are positive scalar multiples of each other. As the second coordinates of the ${\bf a}_i$ are fixed such proportional ${\bf a}_i$ would be given by
$${\bf a}_1:=(\lambda,1),\quad {\bf a}_2:=(2\lambda,2), \quad {\bf a}_3:=(3\lambda,3)\ ,$$
and condition $(2)$ enforces $\lambda={4\over3}$. It follows that there is exactly one point $(x,y,z)\in{\mathbb R}^3$ satisfying the two equations, namely the point $\bigl({4\over3},{8\over3},4\bigr)$.
