For real numbers $x>0, y>0, z>0$ and $x y z=1 .$ Prove that $$ x^{6}+y^{6}+z^{6} \geq x^{5}+y^{5}+z^{5} $$
How to show this?
I tried using https://en.wikipedia.org/wiki/Muirhead%27s_inequality , but without success.
I tried prepare simple form Muirhead, but I don't understand information $xyz=1.$
We need to prove that $$x^6+y^6+z^6\geq(x^5+y^5+z^5)\sqrt[3]{xyz},$$ which is true by Muirhead because $$(6,0,0)\succ\left(\frac{16}{3},\frac{1}{3},\frac{1}{3}\right).$$ Also, we can use AM-GM here: $$\sum_{cyc}x^6=\frac{1}{18}\sum_{cyc}(16x^6+y^6+z^6)\geq\sum_{cyc}\sqrt[18]{\left(x^6\right)^{16}y^6z^6}=\sum_{cyc}\sqrt[18]{x^{90}}=\sum_{cyc}x^5.$$