finding range of function of three variables Three real numbers $x$, $y$, $z$ satisfy the following conditions.
$x^{2}+y^{2}+z^{2}=1~$, $~y+z=1$    
Find the range of $~x^{3}+y^{3}+z^{3}~$ without calculus.
I solved this problem only with Lagrange-Multiplier and wonder if there exist other methods.
 A: Let $z=1-y$ to get $x^2+y^2+(1-y)^2 = x^2 + 2 y^2 -2y +1 = 1$, or $x^2 + 2 y^2 -2y = 0$.
This gives $x = \pm\sqrt{2y(1-y)}$ and $y \in [0,1]$.
Substituting gives:
$x^3+y^3+z^3 = \pm (2y(1-y))^\frac{3}{2} +3 y(y-1) +1$, with $y \in [0,1]$.
Let $\phi_\pm(y) = \pm (2y(1-y))^\frac{3}{2} -3 y(1-y) +1$, with $y \in [0,1]$.
Let $\eta(y) = y(1-y)$, then for $y \in [0,1]$, we have $n(y) \in [0,1]$, $\eta(y)=\eta(1-y)$, $\eta$ is minimized at $y=\pm 1$, and maximized at $y=\frac{1}{2}$, and $\eta(\frac{1}{2}) = \frac{1}{4}$.
Hence $\eta(y) \in [0,\frac{1}{4}]$ for $y \in [0,1]$.
We see that $\phi_\pm(y) = \eta(y) (\pm 2\sqrt{2}\sqrt{\eta(y)}-3) +1$, and note that $(\pm 2\sqrt{2}\sqrt{\eta(y)}-3) \le 0$ for all $y \in [0,1]$.
It follows that both $\phi_+, \phi_-$ have a maximum value of $1$ for $y \in \{0,1\}$. Clearly $\phi_-(y) \le \phi_+(y)$, so to find a minimum value we need only consider $\phi_-$. 
Since $\eta(y) \in [0,1]$, to find the minimum value, we need only consider of $x^2(-2\sqrt{2}x-3)+1$, for $x \in [0,\frac{1}{2}]$. This is easily seen to be decreasing for $ x \ge 0$, hence the minimum value occurs at $x=\frac{1}{2}$. and the minimum value is $\frac{1}{4}-\frac{1}{2 \sqrt{2}}$.
A: to find max value, we only consider $f(u)=1-3u^2+2\sqrt{2}u^3, u=\sqrt{y(1-y)}, 0 \le u \le \dfrac{1}{2}$.we try first to see if it's increasing or decreasing function. $f(0)=1, f(\dfrac{1}{2})=1-\dfrac{3-\sqrt{2}}{4} < 1$, so we guess it is a decreasing function. to prove this, let $ u_1>u_2$
$f(u_1)-f(u_2)=2\sqrt{2}u_1^3-2\sqrt{2}u_2^3-3u_1^2+3u_2^2=(u_1-u_2)(2\sqrt{2}(u_1^2+u_2^2+u_1u_2)-3(u_1+u_2))$
since $u_1-u_2>0$, it remains $(2\sqrt{2}(u_1^2+u_2^2+u_1u_2)-3(u_1+u_2)) <0 \iff 2\sqrt{2}(u_1^2+u_2^2+u_1u_2) < 3(u_1+u_2) \iff 2\sqrt{2}(u_1^2+u_2^2+\dfrac{u_1^2+u_2^2}{2}) <3(u_1+u_2) \iff (u_1-\sqrt{2}u_1^2)+(u_2-\sqrt{2}u_2^2) >0 \iff u_1(1-\sqrt{2}u_1)>0 \iff (u_1\le \dfrac{1}{2})  and   ( 1-\sqrt{2}u_1>0) \implies f(u_1)< f(u_2)$
so the max is $1$ when $u=0 \implies x=y=0$.
