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? 
 A: 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 $x = y = \sqrt{\cfrac{100}{\sqrt[3]{100}}}=\sqrt[3]{100}\,$.
A: Let $x= 2\sqrt{5}$ , $y= 2\sqrt{5}$ $z=5 $.
Then of course $xyz=100$
And $x^2+y^2+z^2=20+20+25=65$.
Now set $x=y$ so $z= \frac{100}{x^2}$ therefore $x^2+y^2+z^2=2x^2+\frac{100}{x^4}$. Now define $$f(x)=2x^2+\frac{100}{x^4}$$
By differentiating we get $$f'(x)=\frac{4x^6-4\cdot100^2}{x^5}$$
so $x=100^{1/3}$ is the minimum of $f$. 
However, we have found that $f(2\sqrt{5})=65$ so $f(100^{1/3})<65$
A: Here’s crude but sound approach.
Consider spheres $x^2+y^2+z^2=r^2$, especially for $r<\sqrt{65}$. You know that when $r=1$, the sphere has no intersection with the surface $xyz=100$, while certainly when $r$ is large, the intersection is nonempty.
Now, the set where the intersection is nonempty is closed, thus of form $[r_0,\infty\rangle$. Let’s find this $r_0$: it tells you when the surfaces $xyz=100$ and $x^2+y^2+z^2=r_0^2$ are tangent. When this happens the gradients of $x^2+y^2+z^2$ and $xyz$ are parallel. For the former, the gradient is $(2x,2y,2z)$, and for the latter, $(yz,xz,xy)$. For these vectors to be parallel, we must have $(x,y,z)=\lambda(yz,xz,xy)$. That is, $\lambda=x/yz=y/xz=z/xy$. From $x/yz=y/xz$ we get $x^2=y^2$, so $x=y$, similarly $y=z$. So the tangency happens when $x=y=z$, at the point $(100^{1/3},100^{1/3},100^{1/3})$, where indeed $3\cdot100^{2/3}<65$.
