Local Mathematical Competitions

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 $k\in {\mathbb{N}}^{\ast }$. 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.