# Alternative solution to a Lagrange Method Optimization Problem

Find extrema of $f(x,y,z)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$ subject to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ by reducing variables and then using the Single Variable Method or by using Geometry instead of using the Lagrange Method.

Just to be sure and to get intuition I used the Lagrange method making the constraint $g(x,y,z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1=0$ and solving $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ from which I got the critical value of $(x^*,y^*,z^*)=(3,3,3)$ which is a saddle point.

Then I tried to reduce variables because I can't think of any geometric argument for this problem. Using the restriction I got $z=\frac{xy}{xy-x-y}$, substituting that in the restriction I got $y=\frac{x}{x-1}$ but when I tried to substitute that into z I get $z=\frac{(x-1)x^2}{(x-1)(x^2-x^2+x-x)}$ but the second part of the denominator is zero. Can someone help?

$$x_1=\frac1x$$ $$y_1=\frac1y$$ $$z_1=\frac1z$$
Then the problem is transformed to, $$f(x_1,y_1,z_1)=x_1^2+y_1^2+z_1^2$$
s.t. $x_1+y_1+z_1=1$
The constraint is a plane passing (0,0,1), (0,1,0), and (1,0,0). The objective function is a sphere with radius r. So the minimum r is when the sphere tangents to the plane or the distance from the origin to the plane that gives you $(\frac13,\frac13,\frac13)$. Thus the maximum point is (3,3,3).