(Proofcheck) Prove that a differential equation has no general solution Good evening!
I am turning to you because I've been working on a question for quite some time and can't seem to find a way to solve it.
Problem:
Let $f : \mathbb{R}^d \to \mathbb{R}^d$ be continuously differentiable. Assume that $\forall x : \, (f(x), x) \geq ||x||^3$ (where $(\cdot, \cdot)$ denotes the usual inner product, prove that there is no solution to: 
\begin{align}
\dot{x} = f(x), \, \, x(0) \ne 0
\end{align}
defined $\forall t \geq 0$.
I am also given the following lemma:
Let $g : \mathbb{R} \to \mathbb{R}$ be continuous, and $\alpha$, $\lambda$ be differentiable functions such that:
\begin{align}
\dot{\alpha} > g(\alpha), \dot{\lambda} = g(\lambda), \exists t_0 \in \mathbb{R}  \text{ with } \alpha(t_0) \geq \lambda(t_0).
\end{align}
Then $\forall t > t_0 : \alpha(t) > \lambda(t)$.
What I have (unsuccessfully?) tried:
It is quite straightforward to prove that $||\dot{x}|| \geq ||x||^2$.
I then decided to restrict $x$ to a line in $\mathbb{R}^d$, say the line spanned by the unit vector $\vec{e_1}$.
Then $x = (x_1, 0, ....)$, and in particular, $y := x_1$ satisfies $|\dot{y}| \geq |y|^2$, and $y(0) = y_0$ where $y_0$ is taken to be the first component of $x_0
Now consider $\mu : \mathbb{R} \to \mathbb{R}$ a function on that restriction which satisfies:
$|\dot{\mu}| = |\mu|^2$, $\mu(0) = y_0$. Then by the above lemma, we should have $|y| > |\mu|$ at all points $t> 0$ (although the above lemma does not consider the absolute value but the function in itself, so I'm a bit skeptical about that bit). But then, the solution $\mu$ will be of the form $t \mapsto (t - 1/y_0)^{-1}$. Now if $y_0 > 0$, then this singularity will happen somewhere on the positive real line. And $|y| > |\mu|$ will imply $y$ also has a singularity $\implies$ it's not defined $\forall t \geq 0$. (note that I only considered the case $y_0 > 0$ because I'm not sure about the case $y_0 < 0. It's just one of many things I'm not sure about)
Now, $f$ being continuously differentiable, it is locally Lipschitz continuous, so by the Picard-Lindelöf theorem, the differential equation has a unique solution in any sufficiently small neighbourhood of any point. And here's the dodgy(iest) bit of the proof:
Assume we have a general solution to the first equation (not only the restriction) that is defined $\forall t \geq 0$. Then if we take a neighbourhood that contains the singularity (which happens somewhere on the line spanned by $\vec{e1}$), then by unicity, both solutions (our general one, and the one we found by restricting ourselves) should coincide. So our general solution also has a singularity.
I hope I made this clear, let me know if I haven't. Thanks for taking the time to read it. If you have any clue on how to make it more rigorous (or any hint to try something else, in case this is completely wrong), I would be really thankful! :)
 A: You seem to be on the right track, but


*

*There is no need to restrict $x$ to a line in $\Bbb R^d$.

*$\Vert \dot x \Vert$ is the norm of the derivative, but you seem to be using it for the derivative of the norm. Similarly at $|\dot{\mu}| = |\mu|^2$.

*It is not a-priory clear that $\Vert  x(t) \Vert$ is never zero, and therefore differentiable everywhere.

*There is no need to apply Picard-Lindelöf in the neighborhood of the “singularity”.


I would proceed as follows: Assume that a solution $x: [0, \infty) \to \Bbb R^d$ of 
$$
\dot x(t) = f(x(t))
$$
with $x(0) \ne 0$ exists. Define
$$
\alpha: [0, \infty) \to \Bbb R, \alpha(t) = \Vert x(t) \Vert^2 = (x(t), x(t)) \, .
$$
Then $\alpha(0) > 0$ and
$$
 \dot \alpha(t) = 2 (x(t), \dot x(t)) = 2 (x(t), f(x(t))) \ge 2 \Vert x(t) \Vert^3 = 2 \alpha(t)^{3/2} \, .
$$
In particular, $\alpha$ is monotonically increasing and therefore strictly positive for all $t \ge 0$. It follows that
$$ 
\dot \alpha(t) > \alpha(t)^{3/2} \, .
$$
Now consider the differential equation
$$
 \dot \lambda(t) = \lambda(t)^{3/2} \, , \, \lambda(0) = \alpha(0) > 0 
$$
which has the solution
$$
 \lambda(t) = \frac{1}{\left( \frac{1}{\sqrt{\alpha(0)}} - \frac t 2\right)^2}
$$
on the interval $0 \le t < \frac{2}{\sqrt{\alpha(0)}} =: T$. Now you can apply the lemma and conclude that
$$
\alpha(t) > \frac{1}{\left( \frac{1}{\sqrt{\alpha(0)}} - \frac t 2\right)^2}
$$
for $0 < t < T$. This is a contradiction, because the right-hand side is unbounded for $t \to T$, whereas the left-hand side has the finite limit $\alpha(T)$.
