<s>[INST] ${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ 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}$.
 [/INST] Writing them out,
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$
becomes
$x(x-1)+(n-x)(n-x-1) \le (n-1)(n-2)
$
or
$x^2-x+n^2-n(x+x+1)+x^2-x
\le n^2-3n+2
$
or
$2x^2-2x
\le n(2x+1-3)+2
$
or
$2x^2-2x
\le n(2x-2)+2
$
or
$x^2-x
\le n(x-1)+1
$
or
$x(x-1)
\le n(x-1)+1
$
or
$(x-n)(x-1)
\le 1
$
which is true for
$x \ge 1$
and
$x \le n-1$.
This is false for $x=0$
or
$x=n$
where this algebra does not work.
</s>