Is it possible to show ${k\choose 2} \ge \sum_{i } {x_i\choose 2} $ such that $\sum_i x_i =k$ I want to know if this problem can be verified or rejected.
${k\choose 2} \ge \sum_i  {x_i\choose 2  }$ such that $\sum_i x_i =k$ and $x_i, \;k\in \mathbb N.$
For example
$2+3=5$  and ${5 \choose 2} \ge  {3\choose 2} + {2\choose 2} $ 
I tried to make a counterexample but I couldn't  find anything, I wanted to prove it using the definition too, but I did not get anything. Is there a way to prove, or  is there a counterexample?

The left-hand side of your inequality counts the distinct pairs from a size-$k$ set, while the right-hand side takes a partition into size-$x_i$ disjoint subsets, then counts the number of pairs that don't cut across partitions.
