${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ In a proof, the author states that it is clear that:
Given $x\geq 1$ and $ n-x \geq 1$ and finally also $n\geq 2$
$${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$$
This is not immediately clear to me. Of course, If I have $n$ objects and I split them up in $x$ and $n-x$ objects and then I choose to form pairs, amongst these 
 two subsets, I end up with fewer pairs than if I would consider the bigger set, so it is certainly smaller than $ n \choose 2$. 
I just don't see how it is smaller than $ {n-1\choose 2}$.

\begin{align}
{x\choose2} +{n-x\choose2} 
 &=\frac{(x-1)x}2 +\frac{(n-x)(n-x-1)}2 \\
 &=\tfrac12 (x^2-x +n^2-nx-n -nx+x^2+x) \\
 &=\tfrac12 (2x^2 -2nx+n^2-n) \\
 &=\tfrac12 (2x(x-n) +n(n-1))
\end{align}
Given $1 \leq x \leq n-1$ (and $x-n \leq -1$), then
\begin{align}
 &\leq \tfrac12 (2(n-1)(-1) +n(n-1)) \\
 &=\frac{(n-1)(n-2)}2 \color{blue}{\cdot \frac{(n-3)!}{(n-3)!}} \\
 &=\frac{(n-1)!}{2!(n-3)!} \\
 &={n-1\choose2}.
\end{align}
