More elegant proofs of $\binom a2+\binom b2\leq \binom{a+b-1}2$

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.

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 ($\color{orange}{a\ge 1},\color{red}{a+1\ge 2}...\color{green}{a+b-2\ge b-1}$)