<s>[INST] There exist $3$ positive numbers $x,y,z$ such that $x\cdot y\cdot z=100$ but $x^2+y^2+z^2<65$. I have to show this using a form of calculus.
I know the answer is the cube root of $100$. I did this using intuition and upper and lower bounds but I have no idea how to show it from a calculus standpoint. Maybe showing the intersection of the surfaces? 
 [/INST] Hint: by AM-GM $\,
x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2 y^2 z^2} = 30 \sqrt[3]{10} \simeq 64.63\,$, with equality iff $x=y=z\,$.

[ EDIT ]   Equivalent proof using Jensen's inequality for the concave function $\ln(x)$ on $\mathbb{R}^+\,$:

$$
\ln \left(\frac{1}{3}x^2+\frac{1}{3}y^2+\frac{1}{3}z^2\right) \ge \frac{1}{3}\left(\ln(x^2)+\ln(y^2)+\ln(z^2)\right) = \ln(\sqrt[3]{x^2y^2z^2})
$$
By monotonicity of $\ln(x)$ on $\mathbb{R}^+$ it follows that $\cfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{x^2y^2z^2} = 10 \sqrt[3]{10}\,$, and because $\ln(x)$ is strictly concave the equality holds iff $x=y=z\,$.

[ EDIT #2 ] Yet another way to find the minimum of $x^2+y^2+z^2$ using "more" calculus.

Starting from $x^2+y^2 \ge 2xy\,$ with equality iff $x=y\,$, it follows that $x^2+y^2+z^2 \ge 2xy + z^2 = \cfrac{200}{z} + z^2\,$ with equality iff $x=y=\sqrt{\cfrac{100}{z}}\,$.
Let $f(z)=z^2+\cfrac{200}{z}\,$ then $f'(z)=2z - \cfrac{200}{z^2} = 0 \iff z^3 = 100\,$, so the minimum is attained for $z= \sqrt[3]{100}\,$, and $x = y = \sqrt{\cfrac{100}{\sqrt[3]{100}}}=\sqrt[3]{100}\,$.
</s>