Show the origin is the only critical point of $f$ $f(x,y,z)=\frac1 2(x^2+y^2+z^2)+a(xy+yz+zx)$
For $a\neq-\frac1 2$ and $a\neq1$, show that the origin is the only critical point of f.
I have found $\nabla f =\begin{pmatrix} x+a(y+z) \\ y+a(x+z) \\ z+a(x+y) \end{pmatrix}  $
Equating this to zero to find critical points, I get $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ = $\begin{pmatrix} -a(y+z) \\ -a(x+z) \\ -a(x+y) \end{pmatrix}$ (1)
Subtracting each row in (1), I get: $x-y = a(x-y), x-z=a(x-z), y-z=a(y-z)$
, therefore $a=1$ gives non-zero critical points.
Then subbing $z$ in (1) into the other two equations, I get:
$x(1-a^2)=y(a^2-a)$ and $y(1-a^2)=x(a^2-a)$, then subbing y into first equation and canceling, I get:
$x(1+a)=\frac{a^2} {1+a}x$, this means that $(1+a)^2=a^2$, solving we get: $a=-\frac12$
Since, $a\neq1$ and $a\neq -\frac12 $, we can have only critical points for when $x,y,z=0$
Does this prove that this is the only set of critical points? Surely it shows that $a=1$ and $a= -\frac12 $ give critical points for $x,y,z$, but not that there are no other critical points. How would I show that there are no other critical points?

Suppose $a\neq-\frac{1}{2}$ and $a\neq 1$. Sum the components of $\nabla f$ gives
$$
(x+y+z)+2a(x+y+z)=0
$$
If $x+y+z\neq 0$, then we would have $a=-\frac{1}{2}$. So $x+y+z=0$. Use this in $x=-a(y+z)$:
$$
x=-a(y+z)=-a(-x)=ax.
$$
If $x\neq 0$, then $a$ would be $1$. So $x=0$. Similarly, you can apply $x+y+z=0$ to $y=-a(x+z)$ to get $y=0$. And with $x=y=0$, $z$ must also be $0$.
So $x=y=z=0$ is the only solution to $\nabla f=0$ when $a\neq-\frac{1}{2}$ and $a\neq 1$.
P.s. Thanks smith for \nabla.
