Question: Let $X\subseteq\mathbb{R}$ with $\left|X\right|=n$ and define $Y=\left\{ x+y\vert x,y\in X\right\}$. Prove that 
$$\left|Y\right|\geq2n-1$$

Answer: 
Let $x\_1<x\_2<\dots<x\_n$ be the sorted list of the elements of $X$. Then
$$
x\_1+x\_1<x\_1+x\_2<x\_1+x\_3<\dots<x\_1+x\_n<x\_2+x\_n<x\_3+x\_n<\dots <x\_n+x\_n
$$
are $2n-1$ distinct members of $Y$, so $|Y|\ge 2n-1$.
Use induction on $n$. Write $X = \{x\_1, \ldots, x\_n\}$, where $i<j$ implies $x\_i < x\_j$. Then by induction 
$Y\_{n-1} =\{x\_i+x\_j; i,j < n\}$ has at least $2(n-1) - 1$ elements. But the two numbers $x\_n +x\_{n-1}$ and $x\_{n} + x\_{n-2}$ are each larger than any elements in $Y\_{n-1}$, and are both in $Y$.