<s>[INST] How to show that $f\left( \frac{ x + y }{ 2 }\right ) \leq \frac{ f( x ) + f( y ) }{ 2 }$ when $f''(x) \geq 0$. I need to show that if $f: (a,b) \to \mathbb{ R }\text{ with}\;\; f''( x ) \geq 0$ for all $x \in (a,b)$, then $f\left( \frac{ x + y }{ 2 } \right) \leq \frac{ f( x ) + f( y ) }{ 2 }$.
I know that since $f''( x ) \geq 0$, then $f'(x)$ is monotone increasing. I'm not really sure where to go from here.
 [/INST] Since $f''(x) \geq 0$, $f'(x)$ is monotone increasing (as you already observed). Let $x, y \in (a, b)$ and WLOG $x < y$. Therefore, $x < \frac{x + y}{2} < y$. Now, applying MVT in $\left(x, \frac{x+y}{2}\right)$ and $\left(\frac{x + y}{2}, y\right)$, we get
$$
f'(c_1) = \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}}\\
f'(c_2) = \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}}
$$
for some $c_1 \in \left(x, \frac{x+y}{2}\right)$ and some $c_2 \in \left(\frac{x + y}{2}, y\right)$.
Now $c_1 \leq c_2$, therefore $f'(c_1) \leq f'(c_2)$. This implies,
$$
\frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}} \leq 
\frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} \\
f\left(\frac{x + y}{2}\right) \leq \frac{f(x) + f(y)}{2}
$$
</s>