Find a maximum of: $x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $

Question

$x,y,z \ge 0 $ , $ x+y+z =1$

Find a maximum of: $$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x $$

and when it is reached.

my attempt:

1) $$x^{2016} \cdot y+y^{2016} \cdot z+z^{2016} \cdot x \le $$ $$\le \left( \frac{2016x+y}{2017}\right)^{2017}+...$$

2) $z=1-x-y$ $$f(x;y)=x^{2016} \cdot y+y^{2016} \cdot (1-x-y)+(1-x-y)^{2016} \cdot x $$

$$\frac{df}{dx}=2016x^{2015}\cdot y-y^{2016}+(1-x-y)^{2016} -2016(1-x-y)^{2015}x $$

$$\frac{df}{dy}=...$$

Answer 1

1. $x\geq y\geq z$

Hence, $x^{2016}y+y^{2016}z+z^{2016}x\leq(x+z)^{2016}y=2016^{2016}\left(\frac{x+z}{2016}\right)^{2016}y\leq$

$\leq2016^{2016}\left(\frac{2016\cdot\frac{x+z}{2016}+y}{2017}\right)^{2017}=\frac{2016^{2016}}{2017^{2017}}$.

2. $x\geq z\geq y$.

In this case $x^{2016}y+y^{2016}z+z^{2016}x\leq x^{2016}z+y^{2016}x+z^{2016}y\leq(x+y)^{2016}z=$

$=2016^{2016}\left(\frac{x+y}{2016}\right)^{2016}z\leq2016^{2016}\left(\frac{2016\cdot\frac{x+y}{2016}+z}{2017}\right)^{2017}=\frac{2016^{2016}}{2017^{2017}}$.