<s>[INST] 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.
 [/INST] 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 $u=0 \implies x=y=0$.
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