The claim plainly holds when $a=b$.

When $a\ne b$, the original claim is equivalent to $$\frac{a^{n+1}-b^{n+1}}{(n+1)(a-b)}\ge\Big(\frac{a+b}{2}\Big)^n$$

Without loss of generality, let's assume $a > b$ and set $x=\frac{a+b}{2}$, $y=\frac{a-b}{2}$. Then both $x$ and $y$ are positive real numbers and $a=x+y$, $b=x-y$. It follows that $$\begin{align*} \frac{a^{n+1}-b^{n+1}}{(n+1)(a-b)} =\ &\frac{1}{2y(n+1)}\left((x+y)^{n+1}-(x-y)^{n+1}\right)\\ \\=\ & \frac{1}{n+1}\left(\binom{n+1}{1} x^n+\binom{n+1}{3}x^{n-2}y^2+\cdots\right)\\ \\\ge\ &\frac{1}{n+1}\binom{n+1}{1}x^n\\ \\=\ & \left(\frac{a+b}{2}\right)^n \end{align*}$$