This question was asked to be proved by Hölder's inequality-
Let $a, b, c$ be positive real numbers. Prove that for all natural numbers $k$, $(k \ge 1)$, the following inequality holds $$ {a^{k+1}\over b^k}+{b^{k+1}\over c^k}+{c^{k+1}\over a^k} \geq {a^k\over b^{k-1}} + {b^k\over c^{k-1}} + {c^k\over a^{k-1}} $$
which has a beauteous evidence by the Hölder's Inequality, though I tried rearrangement-
Writing the LHS-
$$ {a^k\over b^{k-1}} + {b^k\over c^{k-1}} + {c^k\over a^{k-1}} = {a^kb\over b^{k}} + {b^kc\over c^{k}} + {c^ka\over a^{k}} $$
Setting $ x = {a^k \over b^k} $ and $y$ and $z$ similarly, we get
$$ xa+yb+zc \ge xb+yc+za $$
which should be true as WLOG $a\ge b\ge c$ with no certainty for $x,y$ and $z$.
Is this proof correct? If not can I make any corrections? If so then how?
Thanks!
You can not assume that $a\geq b\geq c$ because our inequality is cyclic and not symmetric.
The following reasoning solves your problem. $$\sum_{cyc}\left(\frac{a^{k+1}}{b^k}-\frac{a^k}{b^{k-1}}\right)=\sum_{cyc}\frac{a^k(a-b)}{b^k}=$$ $$=\sum_{cyc}\left(\frac{a^k(a-b)}{b^k}-(a-b)\right)=\sum_{cyc}\frac{(a-b)(a^k-b^k)}{b^k}\geq0.$$