Johan Gunardi wrote: Let $ x,y,z$ be real numbers. Find the minimum value of $ x^2 + y^2 + z^2$ if $ x^3 + y^3 + z^3 - 3xyz = 1$. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231939

Let $ x + y + z = k$, then $ x^2 + y^2 + z^2 - xy - yz - zx = \frac {1}{k}$ and clearly $ k > 0$. Now we obtain \[ x^2 + y^2 + z^2 + 2(xy + yz + zx) = k^2 \] \[ xy + yz + zx = \frac {k^3 - 1}{3k} \] \[ x^2 + y^2 + z^2 = \frac {k^3 + 2}{3k} \ge 1 \] by AM-GM Inequality. Let $ x = y = 0$ dan $ z = 1$, we know that the minimum value is $ 1$.

let $x^2+y^2+z^2=q,x+y+z=p$ then $p(q-\frac{p^2-q}{2}=1$ hence $q=\frac{\frac{2}{p}+p^2}{3}$ by derivatives when $p=1,q_{min}=1$ QED

littletush wrote: let $x^2+y^2+z^2=q,x+y+z=p$ then $p(q-\frac{p^2-q}{2}=1$ hence $q=\frac{\frac{2}{p}+p^2}{3}$ by derivatives when $p=1,q_{min}=1$ QED Actually you don't need derivatives. By AM-GM, $\frac{\frac{2}{p}+p^2}{3}=\frac{1}{3}\left(\frac{1}{p}+\frac{1}{p}+p^2\right)\ge \frac{1}{3}\left(3\sqrt[3]{\frac{1}{p} \frac{1}{p} p^2}\right)=\boxed{1}$. (Equality occurs iff $\frac{1}{p}=p^2\Rightarrow p=1$.)

On the other hand, here is an unfortunate substitution I found which leads to an obvious use of derivatives. By the "difference of three cubes" (this is the official name) factorization we have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=1$. Let $a:=x+y+z$ and $b:=xy+yz+zx$ so that the equation becomes $a(a^2-3b)=1$. We want to minimize $a^2-2b$, so we let $X:=a^2-2b$ to get $a(X-b)=1\Rightarrow X=\frac{1}{a}+\frac{b}{a}$. Now solving $a(a^2-3b)=1$ for $b$ gives $b=\frac{a^2-\frac{1}{a}}{3}$. Hence, substituting into our $X$-equation and simplifying, we obtain $X=\frac{a^2+3a-1}{3a^2}$. Differentiating $X$ w.r.t. $a$ and setting it equal to $0$ to find the minimum yields \begin{align*}X' & =\frac{a^3-3a+2}{3a^3}=0\\ &\Leftrightarrow a^3-3a+2=0\\ &\Leftrightarrow (a-1)^2(a+2)=0.\end{align*} Clearly $a>0$ (which can be seen from $a((x-y)^2+(y-z)^2+(z-x)^2)=1$), hence if there is a minimum it must occur at $a=1$. Now $X''=\frac{2(a-1)}{a^4}$ which is $0$ at $a=1$. So we evaluate $X'$ at $a\pm \epsilon$ to find that the function is increasing at $a=1$. Thus, there exists a minimum at this value of $a$. Finally, plugging $a=1$ into into $X$ yields $\boxed{X=1}$, which is the desired minimum.