Find the maximum and minimum values of the function $f(x,y,z)=(xyz)^2$ where $(x,y,z)$ is on the sphere $x^2+y^2+z^2=r^2$.
Then show using above that $(abc)^{1/3} \leq (a+b+c)/3$. For non-negative $a,b,c$
For the first part I have set up the Lagrangian and calculated the partial derivatives.
That is we have $$2xy^2z^2+2\lambda x =0\\2x^2yz^2+2\lambda y =0\\2x^2y^2z+2\lambda z =0\\x^2+y^2+z^2-r^2=0$$
Now since the point lie on the sphere we have $x,y,z \neq0$ hence $\lambda=-y^2z^2=-x^2z^2=-x^2y^2 \implies x^2=y^2=z^2$ (I think).
So we get $3x^2=r^2 \implies x^2=r^2/3$ and then the maximum and minimum are the same and we have $f(x,y,z)=(xyz)^2=r^6/27$.
Okay so firstly I want to ask am I correct up until here?
And secondly how do I prove the last inequality?
We have that $f(x,y,z) \ge 0$ for all $(x,y,z)$. Now $f(0,r/{\sqrt 2}, r/{\sqrt 2}) = 0$, hence the minimum of $f$ is $0$.
I am not sure what you mean by $x,y,z \neq 0$, but if it is "$ x \neq 0$ and $y \neq 0$ and $z \neq 0$", then it is incorrect. What we know is that at least one of them should be $\neq 0$, but this is not important.
From the system you have, multiply the first equation by $x$, the second by $y$ and the third by $z$ and subtract the equations from each other to get $x^2 = y^2 = z^2$ (realizing that $\lambda = 0$ corresponds to $f(x,y,z) = 0$, and then that we may suppose $\lambda \neq 0$). Then, $x^2 = y^2 = z^2 = r^2/3$. Hence, $f$ achieves its maximum $6$ times (at the points $(\pm r/{\sqrt 3}, \pm r/{\sqrt 3},\pm r/{\sqrt 3})$), and this maximum is $f(r/{\sqrt 3},r/{\sqrt 3},r/{\sqrt 3}) = r^6/27$.
See Wore's comment for how to get the inequality.