$2^n-2$, induct on $n$.

Let $x=\frac{a}{a+b},y=\frac{b}{a+b}$, we want to prove that $\frac{\frac{(a+b)^n-a^n-b^n}{(a+b)^n}}{\frac{ab^n+a^nb}{(a+b)^{n+1}}}\leq 2^n-2$ $\Leftrightarrow (a+b)^{n+1}-(a^n+b^n)(a+b)\leq (2^n-2)(ab^n+a^nb)$ $\Leftrightarrow \sum_{i=1}^{n}{a^ib^{n+1-i}\binom{n+1}{i}} -a^nb-ab^n\leq (2^n-2)(ab^n+a^nb)$ $\Leftrightarrow \frac{1}{2}\sum_{i=1}^{n}{\left( (a^ib^{n+1-i}+a^{n+1-i}b^i)\binom{n+1}{i}\right) } \leq (2^n-1)(ab^n+a^nb)$ Since $\frac{1}{2}\sum_{i=1}^{n}{\binom{n+1}{i}} \leq \frac{1}{2}(2^{n+1}-2)=2^n-1$ It is suffice to prove that $a^ib^{n+1-i}+a^{n+1-i}b^i\geq ab^n+a^nb$ for all $i=1,2,...,n$ Which is equivalent to $(a^{n+1-i}-b^{n+1-i})(a^i-b^i)=(a-b)^2(a^{n-i}+a^{n-i-1}b+...+ab^{n-i-1}+b^{n-i})(a^{i-1}+a^{i-2}b+...+ab^{i-2}+b^{i-1}) \geq 0$ which is obvious.