Prove a inequality property of convex function 
Problem: Let $f: \mathbb{R}^n \to \overline{\mathbb{R}}$ be the convex function. Let $x_1,x_3
\in E$ (Euclidean space in $\mathbb{R}^n$) and $x_2 \in (x_1,x_3)$.
Prove that $$\dfrac{f(x_3) -f(x_2)}{\Vert x_3-x_2 \Vert} \ge\dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1\Vert}.$$

My attempt: Since $x_2 \in (x_1,x_3)$ then there exists $t \in (0,1)$ such that $x_2 = tx_1 + (1-t)x_3$. Thus we have
$$f(x_3) - f(x_2)  \ge f(x_3)-tf(x_1)-(1-t)f(x_3) = tf(x_3)-tf(x_1).$$
Therefore
$$\dfrac{f(x_3)-f(x_2)}{\Vert x_3-x_2\Vert} \ge \dfrac{t}{\vert t \vert}\dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert} \ge -\dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert} = \dfrac{f(x_1)-f(x_3)}{\Vert x_1-x_3\Vert}.$$
I have tried many different ways to have a result like the solution above but I failed. I wonder that the problem is right or not?
 A: Since $x_2 \in (x_1,x_3)$, there exists $t \in (0,1)$ such that $x_2 = tx_1 + (1-t)x_3$. Thus we have
$$f(x_2) = f(tx_1+(1-t)x_3) \le tf(x_1) + (1-t)f(x_3).\ (1)$$
Firstly, from (1) we have
\begin{align*}
  & \dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1 \Vert}  \le \dfrac{(1-t)\left(f(x_3)-f(x_1)\right)}{(1-t)\Vert x_3-x_1\Vert} = \dfrac{f(x_3)-f(x_1)}{\Vert x_3 - x_1\Vert}\\ 
  \text{and }& \dfrac{f(x_3) - f(x_2)}{\Vert x_3-x_2 \Vert} \ge \dfrac{t(f(x_3)-f(x_1)}{t\Vert x_3-x_1\Vert} = \dfrac{f(x_3)-f(x_1)}{\Vert x_3-x_1\Vert}.
 \end{align*}
This yields
$$\dfrac{f(x_3) -f(x_2)}{\Vert x_3-x_2 \Vert} \ge \dfrac{f(x_2)-f(x_1)}{\Vert x_2-x_1\Vert}. (\text{qed}).$$
A: 
Since $x_2 \in (x_1,x_3)$ then there exists $t \in (0,1)$ such that $x_2 = tx_1 + (1-t)x_3$

Writing it as $\,x_2-x_3=t(x_1-x_3)\,$, it follows that $\,\|x_2-x_3\|=|t|\,\|x_1-x_3\|\,$. Given that $\,t\,$ is positive, this means $\,t=\dfrac{\|x_2-x_3\|}{\|x_1-x_3\|}\,$. Then $\,1-t=\dfrac{\|x_1-x_2\|}{\|x_1-x_3\|}\,$ either directly by symmetry, or using the additive property for collinear points $\,\|x_1-x_3\|=\|x_1-x_2\|+\|x_2-x_3\|\,$.
With that, the convex inequality:
$$
\require{cancel}
\begin{align}
&& \color{red}{\left(t + (1-t)\right)} \cdot \color{black}{f(x_2)} &\;\le\; tf(x_1) + (1-t)f(x_3)
\\ &\iff &t\left(f(x_2)-f(x_1)\right) &\;\le\; (1-t)\left(f(x_3)-f(x_2)\right)
\\ &\iff &\frac{\|x_2-x_3\|}{\bcancel{\|x_1-x_3\|}} \left(f(x_2)-f(x_1)\right) &\;\le\; \frac{\|x_1-x_2\|}{\bcancel{\|x_1-x_3\|}} \left(f(x_3)-f(x_2)\right)
\end{align}
$$
