LEMMA. $D_{n,k+1} D_{n,k}^{-1} \le D_{n,1}^2$.
Proof. By simple calculations we get
$$D_{n,k+1} D_{n,k}^{-1} = \frac{x^{2n+2} + y^{2n+2} + xy(x^{n+k} y^{n-k} + x^{n-k} y^{n+k})}{x^{2n} + y^{2n} + x^{n+k} y^{n-k} + x^{n-k} y^{n+k}}$$
and
$$D_{n,1}^2 = \frac{x^{2n+2} + y^{2n+2} + 2(xy)^{n+1}}{x^{2n} + y^{2n} + 2(xy)^n}.$$
Letting $A = x^{2n+2} + y^{2n+2}$, $B = x^{2n} + y^{2n}$, $C_1 = x^{n+k} y^{n-k} + x^{n-k} y^{n+k}$ and $C_2 = 2x^n y^n$, and by clearing denominators the thesis becomes
$$(C_1 - C_2)(A - xyB) \ge 0.$$
But here we have now $C_1 \ge C_2$ by the AM-GM inequality, and $A \ge xyB$ by the rearrangement inequality (or just some trivial factorization). $\square$

Using the LEMMA with a telescoping product we obtain
$$D_{n,n+1} D_{n,1}^{-1} = \prod_{k=1}^{n} \frac{D_{n,k+1}}{D_{n,k}} \le D_{n,1}^{2n}.$$
Substituting, our hypothesis corresponds to $D_{n,n+1} \ge 1$, and therefore
$$D_{n,1}^{2n+1} = \left( \frac{x^{n+1} + y^{n+1}}{x^n + y^n} \right)^{2n+1} \ge D_{n,n+1} \ge 1,$$
whence the thesis.

Alternative Solution.
Suppose $x^{n+1} + y^{n+1} < x^n + y^n$. Then $xy < 1$; otherwise (i.e. for $xy \ge 1$) we would have
$$x^n + y^n \le x^{n+\frac{1}{2}} y^{\frac{1}{2}} + x^{\frac{1}{2}} y^{n+\frac{1}{2}} < x^{n+1} + y^{n+1},$$
absurd. The inequality $x^n + y^n \le x^{n+\frac{1}{2}} y^{\frac{1}{2}} + x^{\frac{1}{2}} y^{n+\frac{1}{2}}$ comes from $1 \le x^{\frac{1}{2}} y^{\frac{1}{2}}$, while the inequality $x^{n+\frac{1}{2}} y^{\frac{1}{2}} + x^{\frac{1}{2}} y^{n+\frac{1}{2}} < x^{n+1} + y^{n+1}$ immediately holds, by writing it $(x^{\frac{1}{2}} - y^{\frac{1}{2}}) \left( x^{n+\frac{1}{2}} - y^{n+\frac{1}{2}} \right) > 0$.
Without loss of generality, we may take $x > 1 > y > 0$ (the other cases are trivial). It follows that $x^k + x^{-k} < y^k + y^{-k}$ for all $k \in \mathbb{N}^*$. This comes from the function $f(t) = t + t^{-1}$ for $t > 0$ being decreasing on $(0, 1]$ and increasing on $[1, +\infty)$, while $x \in (y, y^{-1})$.
From $x^{n+1} + y^{n+1} < x^n + y^n$ follows $x^{n+1} - x^n < y^n - y^{n+1}$.
$$\text{But then } x^{2n+1} - 1 = (x^{n+1} - x^n) \left( 1 + \sum_{k=1}^{n} (x^k + x^{-k}) \right) < (y^n - y^{n+1}) \left( 1 + \sum_{k=1}^{n} (y^k + y^{-k}) \right) = 1 - y^{2n+1},$$
this is to say, in the end, $x^{2n+1} + y^{2n+1} < 2$, contradiction.