I want to find local and global "max", "min" of $f(x_1,x_,x_3) = x_1+4x_2+9x_3$ subject to $1/x_1+1/x_2+1/x_3=1$
I have found already that there are no global "max" and "min", because by setting $x_1=1, x_2=-x_3$ we can reach $+\infty$ or $-\infty$.
Now, I am not sure how I proceed to find local "max" and "min". I did the following:
$x_1$ = $x_3x_2 \over {x_3x_2-x_3-x_2}$ and substited this into objective.
$f$ = $x_3x_2 \over {x_3x_2-x_3-x_2}$$ +4x_2+9x_3$
I took the derivatives with respect to $x_2,$ and $x_3$ and made it equal to $0$. Then I got:
$x_3^2 \over (x_3x_2-x_3-x_2)^2$ $=4$
$x_2^2 \over (x_3x_2-x_3-x_2)^2$ $=9$
I found two solutions $(6,3,2)$ and $(-4, 4/3, 2)$. I checked my answer with wolframalpha
Apparently $(6,3,2)$ is a local minimum and $(-4, 4/3, 2)$ is a saddle point, but how can I find local maximum? According to wolfram it is at (0.535396, 1.62726, -0.674539) even though derivative isn't zero at this point
For positive variables by C-S we obtain: $$x_1+4x_2+9x_3=\left(x_1+4x_2+9x_3\right)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)\geq(1+2+3)^2=36.$$ The equality occurs for $(x_1,x_2,x_3)=(6,3,2),$ which says that we got a minimal value.
The maximal value does not exist.
Take $x_1=x_2\rightarrow+\infty$ and $x_3\rightarrow1.$
If our variables can be negative so the minimal value does not exist.
Take $x_1=x_2\rightarrow-\infty$ and $x_3\rightarrow1$.