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.

The intersection of sphere and plane is a smooth curve, one can apply standard optimization methods.
With the Lagrange functional
$$
L(x,y,z,\lambda,\mu)=x^3+x^3+z^3+λ(x^2+y^2+z^2-1)+μ(x+y+z-1)
$$
one gets the equilibrium conditions 
\begin{align}
3x^2+2λx+μ&=0\\
3y^2+2λy+μ&=0\\
3z^2+2λz+μ&=0
\end{align}
As now $x,y,z$ are solutions of the same quadratic polynomial, two of them have to be equal, wlog. $y=z$. Then from the original equations $x=1-2y$ so that 
$$
(1-2y)^2=x^2=1-2y^2\\
6y^2-4y=0\\
y=z=0\lor y=z=\frac23
$$
which gives $x=0$ in the first case with cubic sum $1$ and in the second case $x=-\frac13$ with cubic sum $\frac{5}{9}$. These are the extremal values that the cubic sum function takes on the circle given by the constraints.
