Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.

Since you got that $x^3+y^3+z^3=1+3xyz$, easy to understand that you meant $x^2+y^2+z^2=1,$ which gives $xy+xz+yz=0.$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=0$ and we obtain:  $$0\leq\prod_{cyc}(x-y)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}w^3-w^6\right),$$ which gives
$$-\frac{4}{27}\leq w^3\leq0$$ and we obtain:
$$x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+3w^3\geq1-\frac{4}{9}=\frac{5}{9}.$$
The equality occurs for example for $(x,y,z)=\left(-\frac{1}{3},\frac{2}{3},\frac{2}{3}\right),$ which says that we got a minimal value.
