$\underline{\textbf{Solution 1}}$

First, by basic AM-GM inequality, we have $$1=\frac{1}{a} + \frac{1}{b} \ge \frac{2}{\sqrt{ab}} \implies \sqrt{ab}\ge 2$$

Then $$\begin{align*} &(a+b)^n-a^n-b^n\\ \\=\ &\binom{n}{1} a^{n-1}b + \binom{n}{2}a^{n-2}b^2 +\cdots +  \binom{n}{n-1}ab^{n-1}\\  \\=\ &\frac{1}{2}\cdot\left(\left(a^{n-1}b+ab^{n-1}\right)\binom{n}{1}+ \left(a^{n-2}b^2+a^2b^{n-2}\right)\binom{n}{2}+\cdots\right)\\ \\ \ge\ & \sqrt{(ab)^n}\left(\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n-1}\right) \\ \\ \ge\ & 2^n(2^n-2)\\ \\=\ &2^{2n}-2^{n+1} \end{align*}$$

$\underline{\textbf{Solution 2}}$

Let $a=1+\frac{1}{t}$, $b=1+t$ where $t$ is a positive real number. Also note that $a+b = ab$. Then $$\begin{align*} & (a+b)^n -a^n-b^n\\ \\=\ & a^nb^n-a^n-b^n\\ \\=\ &(a^n-1)(b^n-1)-1\\ \\=\ &\left(\left(1+\frac{1}{t}\right)^n -1\right)\left((1+t)^n-1\right)-1\\ \\=\ &\left(t^{-1}\binom{n}{1}+t^{-2}\binom{n}{2}+\cdots +t^{-n}\binom{n}{n}\right)\\ & \displaystyle\left(t\binom{n}{1}+c^2\binom{n}{2}+\cdots+t^n\binom{n}{n}\right)-1\\ \\ \ge\ &\left(\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}\right)-1\\ \\=\ & (2^n-1)^2-1\\ \\ =\ &2^{2n}-2^{n+1}\end{align*}$$