https://www.youtube.com/watch?v=Kudlm4ZAQok ### Problem $x,y,z\in\mathbb{R}$ $ \begin{aligned} x+y+z&=1&(1) \\ x^{2}+y^{2}+z^{2}&=1&(2) \\ \min \left(x^{3}+y^{3}+z^{3}\right)&=\ ? \end{aligned} $ #### Solution From the second equation, the minimum value of each term is $0$, therefor the maximum of any term must be 1. This restricts $x,y,z$ each to the interval $[-1,1]$ $ \begin{aligned} 0 &\leq \qquad x^{2}, y^{2}, z^{2} &\leq 1 \\ -1 &\leq \qquad x, y, z &\leq 1 \end{aligned} $ Now square the first equation (1), then subtract the second equation (2) $ \begin{aligned} (x+y+z)^{2}&=1& \text{square }(1) \\ x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z&=1\\ -(x^{2}+y^{2}+z^{2})\qquad\qquad\qquad\qquad &\ -1&\text{subract }(2)\\ 2 x y+2 y z+2 x z&=0\\ x y+ y z+ x z&=0 \end{aligned} $ So at least one of the [[variable|variables]] must be [[Non-Positive]]; the product pairs are either all zeroes, or a mix of positive and negative. By symmetry, allow $x\leq y\leq z$ so that $x$ is the one guaranteed to be non-positive. $x\leq0$ Applying this to the first equation (1), we can write $y$ (or $z$) as the sum of two non-negative parts. $\begin{aligned} x+y+z&=1\\ y&=1-z-x\\ y&=\overset{\text{non-negative}}{\overbrace{(1-z)}}\ \ +\overset{\text{non-negative}}{\overbrace{(-x)}} \end{aligned}$ So $y\geq0$ as well as $z\geq0$. Now to manipulate the sum of cubes $ \begin{aligned} &x^{3}+y^{3}+z^{3} \\ =& x^{3}+(y+z)\left(y^{2}-y z+z^{2}\right) \\ =& x^{3}+(1-x)\left(-\frac{1}{2} y^{2}+\frac{3}{2} y^{2}-\frac{1}{2} z^{2}+\frac{3}{2} z^{2}-y z\right) \\ =& x^{3}+(1-x)\left(-\frac{1}{2}(y+z)^{2}+\frac{3}{2}\left(y^{2}+z^{2}\right)\right) \\ =& x^{3}+(1-x)\left(\frac{1}{2}\right)\left(3\left(1-x^{2}\right)-(1-x)^{2}\right) \\ =& x^{3}+\left(\frac{1-x}{2}\right)\left(3-3 x^{2}-1+2 x-x^2\right)\\ =&x^{3}+\left(\frac{1-x}{2}\right)\left(-4 x^{2}+2 x+2\right) \\ =&x^{3}+(1-x)\left(-2 x^{2}+x+1\right) \\ f(x)=&3 x^{3}-3 x^{2}+1 \end{aligned} $ Differentiate the new function (of only *one* variable) to find where it has relative minimums $ \begin{aligned} \frac{d}{d x}\left(3 x^{3}-3 x^{2}+1\right) &=9 x^{2}-6 x \\ &=\underset{\leq0}{\underbrace{(3 x)}}\underset{\leq0}{\underbrace{(3 x-2)}} \end{aligned} $ Remember that $x\leq0$, and notice the derivative is positive for all of those values, so the minimum of $f(x)$ over $x\in[-1,0]$ occurs when $x=-1$ Using this in the second equation (2) $\begin{aligned} (-1)^2+y^2+z^2&=1\\ y^2+z^2&=0\\ y=0 \quad \text{and}\quad z &=0 \end{aligned}$ And the same fact applied to equation (1) $\begin{aligned} (-1)+y+z&=1\\ y+z&=2\\ \end{aligned}$ This shows that $x$ cannot take the value $-1$ since it would force either one of $y$ or $z$ to be larger than 1, and also simultaneously 0. So back to manipulating the equations again, $ \left\{\begin{array}{rlccl}x+y+z&=1 &\Leftrightarrow &y+z&=1-x \\ x^{2}+y^{2}+z^{2}&=1 &\Leftrightarrow &y^{2}+z^{2}&=1-x^{2}\end{array}\right. $ Squaring the new version of the first equation, $y^2+z^2+2yz=1-x^2-2x$ Then using a modified form of [[AM-GM Inequality]] $y^2+z^2\geq2\sqrt{y^2z^2}$ $y^2+z^2\geq2yz$