$ \color{goldenrod}{\text{Th}}$: Take a continuous map $ [a,b] \overset{f}{\to} \mathbb{R} $, such that $ f’’ $ exists and is $ > 0 $ on $ (a,b) $.
Then $ f((1-t)a+tb) < (1-t)f(a) + tf(b) $ for all $ t \in (0,1) $.
$ \color{goldenrod}{\text{Pf}}$: LHS is just $ f(x) $ evaluated at $ (1-t)a + tb $.
RHS is just “line” $ f(a) + \left( \dfrac{f(b)-f(a)}{b-a} \right)(x-a) $ evaluated at $ (1-t)a + tb $.
So it suffices to show $ \varphi(x) := \left( f(a) + \dfrac{f(b)-f(a)}{b-a} (x-a) \right) - f(x) $ is $ > 0 $ on $ (a,b) $.
$ \varphi’(x) = \dfrac{f(b)-f(a)}{b-a} - f’(x) $, and hence is equal to $ f’(c) - f’(x) $ for some $ c \in (a,b) $. This inturn is $ f’’ (d_x) (c-x) $ for some $ d_x $ between $ c $ and $ x $.
So $\varphi’ $ is $ > 0 $ on $ (a,c) $, $ 0 $ at $ c $, and $ < 0 $ on $ (c,b) $. So $ \varphi $ strictly increases on $ [a,c] $,takes maximum value at $ c $, and strictly decreases on $ [c,b] $.
This, together with $ \varphi(a) = \varphi(b) = 0 $, ensures $ \varphi $ is $ > 0 $ on $ (a,b) $, as needed.