Extension of Definition of Concavity Question
Suppose that the graph of a function $f$ is concave up on an open interval $I$. Show that, for any $a, b \in I$, where $a < b$ and $0 < \lambda < 1$, $$(1 - \lambda)f(a) + \lambda f(b) > f((1 - \lambda)a + \lambda b).$$

My working
From the definition of a function being concave up, $$f(b) - f(a) > (b - a)f'(a).$$
When $b = (1 - \lambda)a + \lambda b$,
$$f((1 - \lambda)a + \lambda b) - f(a) > ((1 - \lambda)a + \lambda b - a)f'(a)$$
$$\implies f((1 - \lambda)a + \lambda b) - f(a) > \lambda (b - a)f'(a)$$

This is where I am currently stuck at. I have a hunch that I am supposed to use the definition once more by substituting another set of values for $b$ and $a$ in order to get rid of the $f'(a)$, but I cannot see what they are. Any intuitions will be greatly appreciated!

A function $f: I \to \Bbb R$ is strictly convex (aka strictly “concave up”) if
$$ \tag 1
f((1 - \lambda)a + \lambda b) < (1 - \lambda)f(a) + \lambda f(b)
$$
for all $a, b \in I$ with $a \ne b$ and all $\lambda \in (0, 1)$. For a differentiable function this is equivalent to
$$ \tag 2
f(y) - f(x) > (y - x)f'(x)
$$
for all $x, y \in I$ with $x \ne y$.
To see that $(2)$ implies $(1)$, set $c = (1 - \lambda)a + \lambda b$. Then
$$
\begin{align}
 f(a)  &> f(c) + (a-c) f'(c) = f(c) + \lambda (a-b) f'(c) \\
 f(b) &> f(c) +(b-c) f'(c) = f(c) +(1-\lambda) (b-a) f'(c) 
\end{align}
$$
and therefore
$$ 
(1-\lambda)f(a) + \lambda f(b) > (1-\lambda)f(c) + \lambda f(c) = f(c) \, .
$$
