Property of a closed path I have the feeling this property holds true but I cannot find a way to prove it:

Let $\gamma: [0,1] \to \mathbb{R}^n$ be a differentiable closed path, i.e. $\gamma(0) = \gamma(1)$. Then there exist two values $\lambda_1 \neq \lambda_2 \in [0,1[$ such that
$$ \left<\gamma'(\lambda_1),  \gamma(\lambda_2) - \gamma(\lambda_1) \right> \geq 0, $$
and
$$ \left<\gamma'(\lambda_2),  \gamma(\lambda_1) - \gamma(\lambda_2) \right> \geq 0. $$

Graphically this means that I can find two points on my path such that the projection of the tangent vector at these points on the line linking these two points face each other.
I would be happy to discuss how this problem could be approached :)
EDIT: I did not want to influence the reflexion in a way that would lead to a dead-end but here is the way I tried to approach the problem.
Let define the function $f: [0,1] \to \mathbb{R}$ as follows:
$$f(\lambda)
=
\left<\gamma'(\lambda),  \gamma(\lambda^{+}) - \gamma(\lambda) \right>
+
\left<\gamma'(\lambda^{+}),  \gamma(\lambda^{+}) - \gamma(\lambda) \right>,
$$
with $\lambda^{+}(\lambda) = mod(\lambda + 0.5, 1)$. Hence, for instance $\lambda^{+}(0.25) = 0.75$ and $\lambda^{+}(0.9) = 0.4$. Obviously $\lambda^{+}(\lambda^{+}(\lambda)) = \lambda$.
We have that $f$ is continuous and $f(\lambda^{+}) = - f(\lambda)$.
Therefore, one can find with the Intermediate Value Theorem a value $\lambda_1$ such that $f(\lambda_1) = 0$. If I define and $\lambda_2 = \lambda^+(\lambda_1)$, this implies that
$$
\left<\gamma'(\lambda_1),  \gamma(\lambda_2) - \gamma(\lambda_1) \right>
+
\left<\gamma'(\lambda_2),  \gamma(\lambda_2) - \gamma(\lambda_1) \right>
= 0 
$$
$$
\Leftrightarrow
\left<\gamma'(\lambda_1),  \gamma(\lambda_2) - \gamma(\lambda_1) \right>
=
\left<\gamma'(\lambda_2),  \gamma(\lambda_1) - \gamma(\lambda_2) \right>.
$$
This means that I can find two points on my curve $\gamma(\lambda_1) $ and $\gamma(\lambda_2)$ such that the projection of the tangent vector at these points on the line between these points either point towards each other, either away from each other.
I am however stuck on how to prove that the former should hold (I believe it should based on intuition and a lot of examples I tried).
 A: What if you want to prove they are orthogonal? Let me rewrite the problem with $\gamma:S^1\to\mathbb{R}^n$, where $S^1$ is the circle $\mathbb{R}/\mathbb{Z}$.
Take $d:S^1\times S^1\to\mathbb{R}$ as $d(x,y)=\|\gamma(x)-\gamma(y)\|^2$, using here the Euclidean norm. This function is continuous from a compact space to $\mathbb{R}$, hence it has a maximum $(\lambda_1,\lambda_2)$. We assume $d(\lambda_1,\lambda_2)>0$ of course.
Then, since this point is a maximum, it is true that $\frac{d}{dx}d(x,\lambda_2)|_{x=\lambda_1}=0$ and $\frac{d}{dy}d(\lambda_1,y)|_{y=\lambda_2}=0$ by differentiability of $\gamma$ (I assume there is no problem at $\lambda=0$ concerning the differentiability).
Then, computing these two derivatives, you obtain exactly
$$\left<\gamma'(\lambda_1),  \gamma(\lambda_2) - \gamma(\lambda_1) \right> = 0,$$
$$\left<\gamma'(\lambda_2),  \gamma(\lambda_1) - \gamma(\lambda_2) \right> = 0.$$
This works perfectly even if if the curve is differentiable but not regular.
In general, you could have equality but not strict positivity, I am thinking about a circular path.
