# 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?

Since $$\dbinom{x}{2}=\frac{x(x-1)}{2}$$, for $$x\ge2$$, your inequality can be written as $$\frac{k(k-1)}{2}\ge \sum_i \frac{x_i(x_i-1)}{2}=\sum_{i}\frac{x_i^2}{2}-\sum_i\frac{x_i}{2}=\sum_{i}\frac{x_i^2}{2}-\frac{k}{2}$$ since, by assumption $$\sum_i x_i=k$$. Also, assume that $$x_i\ge 2$$ for all $$x_i$$ (to avoid trivial complications). So, by rearranging terms, it suffices to show $$k^2\ge \sum_{i}x_i^2$$ or equivalently that $$\left(\sum_i x_i\right)^2\ge \sum_i x_i^2$$
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.
You have to prove $${k^2-k\over 2}\geq {x_1^2-x_1\over 2}+{x_2^2-x_2\over 2}+...+{x_k^2-x_k\over 2}$$
so $${(x_1+x_2+...x_k)^2}\geq x_1^2+x_2^2+...+x_k^2$$
which is obviously true since ali $$x_i>0$$