Find max of $x^7+y^7+z^7$ 
Find max of  $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$

I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck
 A: Here is an elementary way to proceed, which may not be wholly convenient, but which illustrates some interesting techniques.
If $x,y,z$ are the roots of a cubic $t^3-s_1t^2+s_2t-s_3=0$, then we have$$s_1=x+y+z=0$$$$2s_2=2xy+2yz+2zx=(x+y+z)^2-(x^2+y^2+z^2)=-1$$
Thus we have (for some $u$) $$2t^3-t-u=0$$
This gives also (multiply by $t^r$) $$2t^{r+3}-t^{r+1}-ut^r=0$$
Now let $a_r=x^r+y^r+z^r$, and substitute $x, y, z$ successively into the last equation and add the three together to obtain:$$2a_{r+3}-a_{r+1}-ua_r=0 \text{ with }a_0=3, a_1=0, a_2=1$$
Then you can express $a_7$ in terms of $u$ using the recurrence. The constraint on $u$ is that all the roots of the cubic must be real. Combining these two pieces of information you can get a handle on your question.
A: The condition gives $x^2+xy+y^2=\frac{1}{2}$.
Also $$x^7+y^7+z^7=x^7+y^7-(x+y)^7=-7xy(x+y)(x^2+xy+y^2)^2=$$
$$=-\frac{7}{4}xy(x+y)\leq\frac{7}{4}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}=\frac{7}{8\sqrt2}\sqrt{\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}}\leq$$
$$\leq\frac{7}{8\sqrt2}\sqrt{\frac{4}{27}}=\frac{7}{12\sqrt6}$$
because $\frac{x^2y^2(x+y)^2}{8(x^2+xy+y^2)^3}\leq\frac{4}{27}$ it's just $$(x-y)^2(x-z)^2(y-z)^2\geq0.$$
Done!
