Theorem : If $A$ and $B$ are two finite subsets of $\mathbb R$, then
$$\vert A\vert + \vert B\vert -1 \le \vert A+B\vert \le \vert A\vert \vert B\vert$$
and the lower bound is sharp iff $A$ and $B$ are arithmetic progressions of same difference.
Proof : The upper bound is obvious as we can just choose $A$ and $B$ in such a manner that all the sums of the form $a+b$ are different. One example of such a set is
$$A=\{0,1,\dots ,r-1\},\;\; B=r\cdot\{0,1,\dots ,s-1\}$$
so that
$$A+B=\{0,1,\dots , rs-1\}$$
hence completing the proof.
Denoting $A=\{a_1<a_2<\dots <a_r\}$ and $B=\{b_1<b_2<\dots <b_s\}$, and noting that the following chain of inequalities
$$a_1+b_1<a_1+b_2<a_1+b_3<\dots <a_1+b_s<a_2+b_s<\dots <a_r+b_s$$
contains $r+s-1$ different elements of $A+B$ immediately completes the proof of the lower bound.
Now, let us assume that the lower bound is tight. Then, the following chain of inequalities
$$a_1+b_1<a_2+b_1<a_2+b_2<a_2+b_3<\dots <a_2+b_s<a_3+b_s<\dots <a_r+b_s$$
and the previous one must be the same.
The first and the last $r-1$ terms are clearly equal, the other $s-1$ terms determine the equations
\begin{align*}
a_1+b_2&=a_2+b_1\\
a_1+b_3&=a_2+b_2\\
&.\\
&.\\
&.\\
a_1+b_s&=a_2+b_{s-1}
\end{align*}
which on rearranging gives
\begin{align*}
b_2-b_1&=a_2-a_1\\
b_3-b_2&=a_2-a_1\\
&.\\
&.\\
&.\\
b_s-b_{s-1}&=a_2-a_1
\end{align*}
and hence $B$ is an arithmetic progression with difference $a_2-a_1$. Since the roles of $A$ and $B$ are symmetric in the statement, then so must be $A$.
In fact, a more general statement is true.
Theorem : If $A_1,A_2,\dots ,A_k$, $k\ge 2$ are finite subsets of $\mathbb R$, then
$$\vert A_1\vert + \vert A_2\vert +\dots \vert A_k\vert -k+1 \le \vert A_1+A_2+\dots A_k\vert \le \vert A_1\vert \vert A_2\vert\dots \vert A_k\vert$$
and the lower bound is sharp iff $A_1,A_2,\dots ,A_k$ are arithmetic progressions of same difference.
Proof : We proceed by Induction.
The case $k=2$ is precisely the last theorem.
Now, let us assume that the statement is true for $k-1$.
Let $A=A_1+A_2+\dots +A_{k-1}$ and $B=A_k$ so that $A+B=A_1+A_2+\dots +A_{k}$. Using the previous theorem and the Induction Hypothesis, we get
\begin{align*}
\vert A_1\vert + \vert A_2\vert +\dots \vert A_{k-1}\vert -k+2 +\vert A_k\vert -1
&\le \vert A\vert + \vert B\vert -1\\
&\le \vert A+B\vert\\
&\le \vert A\vert \vert B\vert\\
&\le \vert A_1\vert \vert A_2\vert\dots \vert A_k\vert
\end{align*}
and hence the proof follows.
Now, we just notice that the left inequality can be an equality only if
$$\vert A_1\vert + \vert A_2\vert +\dots \vert A_{k-1}\vert -k+2 = \vert A_1+A_2+\dots +A_{k-1} \vert = \vert A \vert$$
and
$$\vert A\vert + \vert B\vert -1 = \vert A+B\vert.$$
The first relation implies that $A_1,\dots ,A_{k-1}$ and hence $A_1+\dots +A_{k-1}=A$ are arithmetic progressions of same difference, and the second relation implies that $B$ is also an arithmetic progression of same difference. This completes the proof.
Corollary : Let $A$ be a finite subset of $\mathbb R$. Then, for an integer $k\ge 2$, we have
$$k\vert A \vert -k+1\le \vert kA \vert\le \vert A \vert^k$$
and the lower bound is sharp iff $A$ is an arithmetic progression.
However, it should be noted that the upper bound in the last corollary is not sharp. The sharp upper bound is given by the following theorem.
Theorem : Let $A$ be a finite subset of $\mathbb R$. Then,
$$\vert kA \vert\le \binom{\vert A \vert +k-1}{k}$$
for any integer $k\ge 1$.
Proof : The number of ways of choosing unordered $k$-tuples from a set of cardinality $\vert A \vert$ is the number of non-zero integer solutions of the equation
$$x_1+\dots +x_{\vert A \vert}=k$$
which is $\binom{\vert A \vert+k-1}{k}$.
This upper bound is indeed sharp and the case $k=2$ of the sharp upper bound is called a Sidon set.
For more on this topic, one can see this and this.
