Procced by induction on $n$. We can check the base cases for $n=1,2,3$. They are all equal and left as an exercise for the reader. Now assume that holds $$(a+b)^n-a^n-b^n \ge \frac{2^n-2}{2^{n-2}} \cdot ab(a+b)^{n-2}$$for some $n \geq 3$ then we will prove that $$(a+b)^{n+1}-a^{n+1}-b^{n+1} \ge \frac{2^{n+1}-2}{2^{{n-1}}} \cdot ab(a+b)^{n-}$$holds as well. Now multiply the statement for $n$ by $a+b$ and we get $$(a+b)^{n+1}-a^{n+1}-b^{n+1}-a b\left(a^{n-1}+b^{n-1}\right) \geqslant \frac{2^{n}-2}{2^{n-2}} \cdot a b(a+b)^{n-1}.$$And then we only need to prove $$ab\left(a^{n-1}+b^{n-1}\right)+\frac{2^{n}-2}{2^{n-2}} \cdot a b(a+b)^{n-1} \geqslant \frac{2^{n+1}-2}{2^{n-1}} \cdot a b(a+b)^{n-1}.$$Now just subtract $\frac{2^{n}-2}{2^{n-2}} \cdot a b(a+b)^{n-1}$ to get $$a b\left(a^{n-1}+b^{n-1}\right) \geqslant\left(\frac{2^{n+1}-2}{2^{n-1}}-\frac{2^{n}-2}{2^{n-2}}\right) \cdot a b(a+b)^{n-1}=\frac{a b(a+b)^{n-1}}{2^{n-2}}.$$And then divide by $2ab$ and we get, after taking the $(n-1)^{\textrm{th}}$ root, the power mean $$\sqrt[n-1]{\frac{a^{n-1}+b^{n-1}}{2}} \geq \frac{a+b}{2}.$$And we are done.