# Why constraint optimization problem becomes infeasible when coefficient is zero

First of all thank you for your time and help. I am more on programming side and have average knowledge of calculus so less formal on mathematical notation. I have encountered a problem in non linear equality constraint optimization

The objective function which has to be minimized $$ax_1^2 + bx_2^2 + cx_3^2$$

subject to constraint

$$\frac {1}{x_1} + \frac {1}{x_2} + \frac {1}{x_3} = T$$

So the Lagrangian is

$$L(..)=ax_1^2 + bx_2^2 + cx_3^2 + \lambda (\frac {1}{x_1} + \frac {1}{x_2} + \frac {1}{x_3} -T)$$

at times the coefficient c takes the value c = o, which results in

$$ax_1^2 + bx_2^2$$

subject to constraint

$$\frac {1}{x_1} + \frac {1}{x_2} + \frac {1}{x_3} = T$$

So the new Lagrangian is

$$L_1(..)=ax_1^2 + bx_2^2 + \lambda (\frac {1}{x_1} + \frac {1}{x_2} + \frac {1}{x_3} -T)$$

Question: My understanding is that $L_1$ has no valid analytical solution nor numerical solution. I am looking for concrete theoretical reasons to convince my senior that in case of c=0 the optimization problem is no longer valid minimization problem. Preferably independent of $L_1$, considering only the objective function and constraint. I will appreciate the multiple explanations from the perspective of geometry, calculus e.t.c.

Let's suppose that the objective function attains a minimum on $x_1,x_2,x_3>0$. Say that minimum is $M$, achieved at $(x_1, x_2, x_3)$.

Then $(x_1, x_2, x_3)$ lies on the surface $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = T$ and also on $a x_1^2 + b x_2^2 + c x_3^2 = M$.

That latter surface is an ellipsoid when $c \not = 0$; the former surface is like so. (The first image is viewed from afar; the second is looking directly down on the point $(1,1,1)$ from the origin.)  We fixed $M$ to be a specific value; now let it vary. It's plausible (indeed, obvious!) that if we make the ellipsoid too small (decreasing $M$ sufficiently) then it will stop intersecting with that surface eventually: it'll start fitting below the bend you see in the picture. So actually there is a minimum of the objective function.

But when $c = 0$, the objective function is now plotting a cylindrical prism rather than an ellipsoid, and now decreasing $M$ just makes the cylinder more squashed. The cylinder will still extend all the way up to infinity, and eventually it'll meet the ever-so-gradually-curving surface, no matter how small $M$ is. So whatever value of $M$ we picked, we could decrease $M$ and still find a solution to the system.