Let $I \subseteq \mathbb{R}$ be an interval. Then a function $f:I \rightarrow \mathbb{R}$ is convex on the interval $I \subseteq \mathbb{R}$ when:
$$\forall a,b \in I, a < x < b \Rightarrow f(x) \le f(a) + \dfrac{f(b)-f(a)}{b-a}.(x-a)$$
Also I know that a function $f:I \rightarrow \mathbb{R}$ is convex on the interval $I \subseteq \mathbb{R}$ iff given $a,b \in I,\ a < b$, the following happens:
$$f(t.a + (1-t).b) \le t.f(a) + (1-t).f(b)\ \ \ \ \forall t \in [0,1]$$
There is another equivalence: A function $f$ is convex on the interval $I \subseteq \mathbb{R}$ iff given $a_1,a_2,...,a_n \in I$ and $t_1,t_2,...,t_n \in [0,1]$ that $\sum t_i = 1$ we have:
$$f(\sum t_i a_i) \le \sum t_i f(a_i)$$
How to prove this last equivalence?