Prove $f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) = 0$ if and only if $x=0$ Properties
Given a $f:\mathbb R^n\to \mathbb R^n$. The function $f$ has the following properties

*

*$f(x) = 0 $ if and only if $x=0$.

*$\nabla f(x)$ are negative semidefine for any $x$.

*$f(x)$ is continuously differentiable

Problem
Is the following iif statement hold?
$$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) = 0 \Leftrightarrow x=0.$$
From right to left, it is straightforward. I am trying to prove the left to right.
Some tries
According to the negative semidefinite of $\nabla f(x)$, we have $\nabla^T f(x)+\nabla f(x)$ are also negative semidefinite, such that
the first part
$$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x) \leq 0 .$$
Again using $\nabla f(x)$ are negative semidefine, by
lemma: If a mapping $f$ is continuously differentiable then $f$ is monotone if and only if its Jacobian matrix $\nabla f(x)$ is positive semidefinite
, we have the second part
$$x\cdot f(x)\leq0.$$
So far, we have
$$f^T(x)(\nabla^T f(x)+\nabla f(x))f(x)+x\cdot f(x) \leq 0.$$
This is the best I can do, I don't know where to go. Or maybe the iif statement is not true in the first place.
 A: After an embarrassingly long time, I can now give my own proof of this problem. I will prove the result by contradiction.
Step 1: Let a point $x_3 \neq 0$, which can satisfy the statement $$f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3)+x_3\cdot f(x_3) = 0 .$$
And also since $x_3\neq0$, $f(x_3) \neq 0$ by property 1.
Step 2: Rewrite the statement as
$$x_3\cdot f(x_3) = -f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3).$$
Since $\nabla^T f(x_3)+\nabla f(x_3)$ is a negative semidefinite matrix, then $-(\nabla^T f(x_3)+\nabla f(x_3))$ is a positive semidefinite matrix, we have
$$x_3\cdot f(x_3) = -f^T(x_3)(\nabla^T f(x_3)+\nabla f(x_3))f(x_3)\geq 0$$
$$x_3\cdot f(x_3) \geq 0$$
Step 3: By the assumption $x_3$ and $f(x)$ are both not $0$, then
$$x_3\cdot f(x_3) > 0$$
Step 4: By $\nabla f(x_3)$ being negative semidefinite, and using the lemma described above, we should have
$$x_3\cdot f(x_3) \leq 0.$$
This is contradicting with the results of Step 3. That is, the point $x_3 \neq 0$ satisfying the statement would not exist.
