Let $F : \mathbb{R^n} \rightarrow \mathbb{R^n}$ be a differentiable function, with positive definite Jacobian matrix. Prove that function $f(x) = \langle x, F(x) \rangle$, with assumptions $x \ge 0, F(x) \ge 0$ has minimum in point $x^*$ if and only if $\langle x^*, F(x^*)\rangle = 0$
I managed to prove the simpler implication: if
$\langle x^*, F(x^*)\rangle = 0, x \ge 0, F(x) \ge 0 \Rightarrow \langle x, F(x) \rangle \ge 0 = \langle x^*, F(x^*) \rangle,$ thus this part is proved, because $F$ is positive definite. How do I even start the other part? The book suggests to use the theorem of F. John. Thank you in advance. $$Edit:$$ The solution book shows is:
If the problem has a solution $x^*$ then, by theorem of F. John, there exist a number $u_0$ and vectors $u^1, u^2$ for which: $$u_0 F(x^*) + u_0 \frac{\partial F}{\partial x}(x^*) - \frac{\partial F}{\partial x}(x^*) u^1 - u^2 = 0, \langle u^1, F(x^*) \rangle = 0, \langle u^2, x \rangle = 0, (u_0, u^1, u^2) > 0. $$ {First, should it be $x^*$ where it states just $x$?}
Now, after multiplication of the first equation by vector $u_0 x - u^1$ we get: $$\langle \frac{\partial F}{\partial x}(x^*)(u_0 x - u^1), u_0 x - u^1 \rangle + u_0^2 \langle F(x^*), x^* \rangle + \langle u^1, u^2 \rangle = 0. $$ {Secondly, how does one obtaim this after the scalar multiplication?}
All addends, by assumptions, are non-negative, so they all equal 0. Thus, $u_0^2 \langle F(x^*), x^* \rangle = 0.$ If $u_0 = 0$, then because of $\langle u^1, u^2 \rangle = 0$ and $u^1 \ge 0, u^2 \ge 0,$ we have $(u_0, u^1, u^2) = 0,$ which is impossible. Hence, $u_0 \ne 0$, so $\langle x^*, F(x^*) \rangle = 0.$ This proves the first implication. The second implication is trivial.
Consider the behavior of $f(x+hx)$ as $h\to 0$: $$ f(x+hx) = \langle x + hx, F(x+hx)\rangle = f(x) + h (\langle x, F(x)\rangle + \langle x, J_F(x)x\rangle) + o(h) $$ where $J_F(x)$ is the Jacobian matrix of $F$ at $x$.
Suppose $x^*$ is a point of minimum. Whether it is interior or boundary point of the domain, $x^*+hx^*$ belongs to the domain for $h>0$. Therefore, the coefficient of $h$ in the above expansion must be $\le 0$. On the other hand, $$\langle x^*, J_F(x^*)x^*\rangle \ge 0$$ (note that being positive semidefinite is enough here). Hence $\langle x^*, F(x^*)\rangle \le 0$. But as you noted, $\langle x^*, F(x^*)\rangle \ge 0$ always holds. Thus $\langle x^*, F(x^*)\rangle = 0$.