If x,y,z are positive real numbers, and $\frac{x^{19}}{y^{19}}+\frac{y^{19}}{z^{19}}+\frac{z^{19}}{x^{19}}≤\frac{x^{19}}{z^{19}}+\frac{y^{19}}{x^{19}}+\frac{z^{19}}{y^{19}}$
How do you prove:
$\frac{x^{20}}{y^{20}}+\frac{y^{20}}{z^{20}}+\frac{z^{20}}{x^{20}}≤\frac{x^{20}}{z^{20}}+\frac{y^{20}}{x^{20}}+\frac{z^{20}}{y^{20}}$
I was able to prove it with a simpler case, however, I am assuming that the first power inequality is true. I don't know if that is correct, but it's just to show a method:
Assuming: $\frac{x^{}}{y^{}}+\frac{y^{}}{z^{}}+\frac{z^{}}{x^{}}≤\frac{x^{}}{z^{}}+\frac{y^{}}{x^{}}+\frac{z^{}}{y^{}}$
Prove: $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}}≤\frac{x^{2}}{z^{2}}+\frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}$
In this case, you would multiply the first power inequality by itself, and you would get: $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}} + 2(\frac{x^{}}{z^{}}+\frac{y^{}}{x^{}}+\frac{z^{}}{y^{}})≤\frac{x^{2}}{z^{2}}+\frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}+ 2(\frac{x^{}}{y^{}}+\frac{y^{}}{z^{}}+\frac{z^{}}{x^{}})$
This would be easy to prove. the first power inequalty is doubled and has flipped sides. Since the terms have flipped sides, in order to keep the new inequality true, $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}}≤\frac{x^{2}}{z^{2}}+\frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}$ must be true.
$$\sum_{cyc}\frac{x^{19}}{z^{19}}\geq\sum_{cyc}\frac{x^{19}}{y^{19}}$$ it's $$(x^{19}-y^{19})(x^{19}-z^{19})(y^{19}-z^{19})\geq0$$ or $$(x-y)(x-z)(y-z)\geq0.$$ Can you end it now?