Looking for a proof for this counting problem Consider all the coefficients in the expansion of 
$$(x^a+x^b)^n$$
where $a,b,n$ are non negative integers.
Claim: The $\color{Red}{\textit{exponent}}$ with the $\color{blue}{\textit {maximum coefficient value}}$ in the expansion is $\color{red}{\lfloor\frac{n(a+b)}2\rfloor}$
Example :
$(x^1+x^3)^4 = \sum\limits_{k=0}^4\binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+\color{blue}{6}x^{\color{red}{8}}+4x^{10}+x^{12}$ 
$\color{red}{\dfrac{4(1+3)}{2} = 8}$.  
I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?

If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $n\choose k$, where $k$ is the largest integer less than or equal to $\frac{n}{2}$. To prove this, consider the ratios $\frac{n \choose k+1}{n\choose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result. 
PS: Note that your exponents must actually appear in the expansion for this to happen.
