Prove that $ f(a+b) \ge f(a)+f(b)$. Let $f$ be a function such that $f^{\prime \prime} (x) \ge 0$ , $f(0)=0$, for each $x \ge 0$.
Prove that if $a,b \ge 0$ , then:
$$ f(a+b) \ge f(a)+f(b)$$ 
It's really hard for me to get an intuition for questions like this, can someone give me a hint and explain in what way did he think to get such intuition? 
 A: For intuition, consider the function $x\mapsto x^2$. If $a, b\geqslant 0$ then it is clear that
$$(a+b)^2 = a^2 + 2ab +b^2 \geqslant a^2+b^2. $$
In general, if $f''\geqslant 0$ on $[0,\infty)$, it follows that $f$ is convex on $[0,\infty)$, and so for any $x\geqslant0$ and $t\in[0,1]$
\begin{align}
f(tx) &= f(tx + (1-t)0)\\
&\leqslant tf(x) + (1-t)f(0)\\
&= tf(x).
\end{align}
Hence if $a,b\geqslant 0$, 
\begin{align}
f(a) + f(b) &= f\left((a+b)\frac a{a+b}\right) + f\left((a+b)\frac b{a+b}\right)\\
&\leqslant \frac a{a+b}f(a+b) + \frac b{a+b}f(a+b)\\
&=f(a+b).
\end{align}
A: Without loss of generality, we assume $0\le a<b$.  Then, from the Mean-Value Theorem, we can write
$$f(a+b)=f(b)+f'(\xi)a \tag 1$$
for $b<\xi < a+b$, and
$$\begin{align}
f(a)&=f(0)+f'(\eta)a\\\\
&=f'(\eta)a  \tag 2
\end{align}$$
for $0 < \eta <a$.  
Next, subtracting $(2)$ from $(1)$ reveals that
$$f(a+b)-f(a)-f(b)=(f'(\xi)-f'(\eta))a \tag 3$$
Now, since $f''(x)\ge 0$ for all $x\in[0,b]$, then $f'(\xi)\ge f'(\eta)$.  Therefore, we have from $(3)$
$$f(a+b)-f(a)-f(b)=(f'(\xi)-f'(\eta))a\ge 0\implies f(a+b)\ge f(a)+f(b)$$
as was to be shown!
A: Hint: If $f''(x) \geq 0 $ then $f'$ is non decreasing. Then use the Mean Value Theorem, to find $z \in (a,x)$ such that $$f(x) = f(a) + f'(z) (x-a)$$
Notice that $f'(z) \geq f(a)$.
