More elegant proofs of $\binom a2+\binom b2\leq \binom{a+b-1}2$ I came to know about the inequality
$$\binom a2+\binom b2\leq \binom{a+b-1}2$$
and tried to prove it.
It was quite easy to derive it using brute force algebraic calculations. All boils down to showing
$$a(a-1)+b(b-1)\leq (a+b-1)(a+b-2)$$
about which you can see here.
But, I want to know whether we can come up with a combinatorial or an intuitive or a geometric explanation. In other words, I am looking for more elegant arguments, or more beautiful proofs of this inequality.
 A: Not sure if this is what you are looking for ....
We  simply have to prove the case when $a\ge 1$ and $b\ge 1$ .Note that the term $\binom{n}{2}$ is basically the sum of the first $n-1$ naturals
Thus our inequality becomes equivalent to $$\color{blue}{1+2+..+a-1}+1+2+..b-1\le \color{blue}{1+2+..+a-1}+a+a+1+..a+b-2$$ or $$\iff \color{orange}{1}+\color{red}{2}+..\color{green}{b-1}\le \color{orange}{a}+\color{red}{a+1}+..\color{green}{a+b-2}$$ which is obvious because for each term in LHS there is a corresponding term in RHS which is greater than or equal to  the term in LHS
($\color{orange}{a\ge 1},\color{red}{a+1\ge 2}...\color{green}{a+b-2\ge b-1}$)
A: In fact, you have
$$
\binom{a+b-1}{2}=\binom{a}{2}+\binom{b}{2}+(a-1)(b-1).
$$
A combinatorial interpretation of this is that you have a set $A$ of size $a$ and a set $B$ of size $b$ with exactly one element in common. If two elements are chosen from $A\cup B$ you get either two elements of $A$ or two elements of $B$, and this includes all cases where the common element is chosen, or else you get one element of $A$ and one element of $B$, neither of which is the common element.
