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 $x^3 + y^3 + z^3$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

You can use Cauchy-Schwarz inequality. $X+y=1-z$, thus $(1-z)^2 = (x+y)^2$ is less than or equal to $(1+1)(x^2+y^2)=2 (1-z^2)$. I get $z$ more than or equal to $-1/3$. $x^3+y^3+z^3= (x^3+y^3)+z^3=(x+y)(x^2-xy+y^2)+z^3= (3(x^2+y^2)-(x+y)^2)/2 + z^3$ doing the necessary substitutions I get $1-3 z^2 +3 z^3$, equals $5/9$.