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. 

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.
