Let $x_{i},i=1,2,\cdots,n$ be real numbers,and such $$|x_{i}-x_{j}|>1(\forall i\neq j)$$ define set $A=\{x_{i}x_{j}+x_{k}|1\le i,j,k\le n\}$,show that $$|A|\ge\dfrac{n(n-1)}{2}$$
How can I go about solving this? I think there is a specific equation for a pigeonhole principle.
On
You could try induction, with the caveat that I'd suggest redefining $A$ with an index, $A_n = \left\{x_{i}x_{j}+x_{k}\space \big\vert \space 1 \le i,j,k\le n\right\}$. The base case ($n = 2$) should give you two numbers, $x_1,x_2$ such that $\left|x_1-x_2\right|>1$ and by hand you can calculate $A_1$. You should easily be able to show that $\left|A_1\right| \geq \frac{2(2-1)}{2} = 1$. So proceed by induction, supposing that $\left|A_n \right| \geq \frac{n(n-1)}{2}$ for all $n \in \{1,2,\ldots, p\}$. The induction hypothesis gives you $\left|A_p \right| \geq \frac{p(p-1)}{2}$ and ultimately you will want to show that $\left|A_{p+1} \right| \geq \frac{p(p+1)}{2}$. As a hint it might help to notice that $\frac{p(p+1)}{2} =\frac{p(p-1)}{2} +p$ and that $A_n \subset A_{n+1}$. Can you prove that $A_{p+1}$ must have at least $p$ more elements in it than $A_p$? If so, you can conclude that $$\frac{p(p+1)}{2} \leq \left|A_p \right|+p \leq \left|A_{p+1} \right|$$
Proof by induction: Suppose $n=3$. Then $A$ has at least 3 elements: $x_1x_2+x_3$, $x_1x_3+x_2$ and $x_2x_3+x_1$. This case holds because $3(2)/2 = 3$.
Now assume that the inequality holds for $n$, we show it holds for $n+1$. Any element in $A_{n}$ is in $A_{n+1}$, so $A_{n}$ has at least $r=\frac{n(n-1)}{2}$ elements. We need to show that it has at least $s=\frac{(n+1)(n-1+1)}{2}=\frac{n(n+1)}{2}$ elements. How many new elements do we need? Taking the difference, $s-r=\frac{n^2+n-(n^2-n)}{2}=\frac{2n}{2}=n$.
How can we construct $n$ new elements in $A_{n+1}$? We could actually construct many more but we can find $n-1$ of the pattern $x_1*x_i+x_{n+1}, 1<i≤n$ and 1 more. $x_2*x_3+x_{n+1}$ for a total of $n$.
So, by induction, $A_{n}$ has at least $\frac{n(n-1)}{2}$ elements.