If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$ 
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
 A: Note the beautiful identity $$y^2+z^2+t^2=\frac{1}{3}\left({(y+z+t)}^2+{(y-z)}^2+{(z-t)}^2+{(t-y)}^2\right)$$  thus we have $$x^2+\frac{1}{3}{(y+z+t)}^2+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=x(y+z+t)$$ $${\left (x-\frac{(y+z+t)}{2}\right)}^2+ \frac{{(y+z+t)}^2}{12}+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=0$$  which means $x=y=z=t=0$
(As its a sum of squares each term must be zero)
A: hint: $x^2+y^2+z^2+t^2=(\frac{x^2}{3}+y^2)+(\frac{x^2}{3}+z^2)+(\frac{x^2}{3}+t^2)
=(\frac{x}{\sqrt{3}}-y)^2+(\frac{x}{\sqrt{3}}-z)^2+(\frac{x}{\sqrt{3}}-t)^2+\frac{2}{\sqrt{3}}\cdot x(y+z+t)\geqslant 0+1\cdot x(y+z+t)$
When would the equality hold?
A: less beautiful
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  & 1 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  & 1 & 0 \\ 
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
2 & 0 & 0 & 0 \\ 
0 &  \frac{ 3 }{ 2 }  & 0 & 0 \\ 
0 & 0 &  \frac{ 4 }{ 3 }  & 0 \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  \\ 
0 & 1 &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 3 }  \\ 
0 & 0 & 1 &  -  \frac{ 1 }{ 2 }  \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
2 &  - 1 &  - 1 &  - 1 \\ 
 - 1 & 2 & 0 & 0 \\ 
 - 1 & 0 & 2 & 0 \\ 
 - 1 & 0 & 0 & 2 \\ 
\end{array}
\right) 
$$
Back substitution says $t=0$  then $z=0$ then $y=0$ then $x=0.$   All that is going on is that the matrix I am calling $H$ is positive definite. With all real variables, each row in my $Q$ must indicate a zero vector ( when multiplied by $(x,y,z,t)^T \; \; \; \; $ )
A: We have$$x^2+y^2+z^2+t^2=x(y+z+t)\iff x^2-\frac34x^2+(y-\frac x2)^2+(z-\frac x2)^2+(t-\frac x2)^2=0$$ i.e.$$\left(\frac x2\right)^2+\left(y-\frac x2\right)^2+\left(z-\frac x2\right)^2+\left(t-\frac x2\right)^2=0$$
We are done
