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? 

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 $f(100^{1/3})<65$
