Ideas on how to proceed with a proof?

Question

Sorry if this is a nonspecific question - I can provide more details but at this point I need general ideas on a proof strategy.

So I recently reduced a rather difficult optimization problem to another problem whether a smooth function $f$ of three real variables $x,y,z$ has a solution for $x,y,z\geq0$, that is $$ f(x,y,z)=0. $$ Unfortunately, the function is a complicated and long algebraic expression containing square roots. From its form it is not obvious that the only non negative solution is whenever $y\geq0,z\geq0$ and $x=0$ which is the desired result and it is also supported by numerical evidence.

My first question is: What are the options now? What would your general strategy be? The standard calculus acrobatics like taking derivatives or Taylor expansion leads to even more complicated expressions. It seems to me that by reducing an optimization problem to a question whether a function is positive/negative is a great simplification.

I tried to set $y=y_0>0,z=z_0>0$ to see what expression I get and it is an algebraic equation with five roots (two complex, two negative and zero). The expression can be transformed into a polynomial equation but this involves squaring (obviously leading to more solutions than there is to the original eqaution $f(x,y_0,z_0)=0$) and moreover it is of degree $>4$ so no explicit solutions anyway.

My second question is: If nothing else, could the proof be concluded at this point? It is not the level of rigor I prefer since the result is not obvious by staring at $f$. On the other hand there must be a number of problems leading to explicitly ''unsolvable'' constraints. In a similar vein, some problems in calculus are solved up to an integral solution since the integral is not expressible in terms of elementary functions (this is a bit lame comparision I admit).

Answer 1

You want to prove that if $f(x,y,z)=0$ and $x,y,z\geq 0$, then $x=0$. As I understand it, you have already proven this statement:

For any value of $y_0,z_0\geq 0$, the function $g(x) = f(x,y_0,z_0)$ has two negative, two complex roots and one root at $x=0$.

Now, simply take any tuple $(x,y,z)$ such that $x,y,z\geq 0$ and $f(x,y,z) = 0$. Now construct the function $g(t) = f(t,y,z)$. Since you know that $g(x) = f(x,y,z)=0$, you know that $x$ is a root of the function $g$, but since all roots of this function are either complex, negative or equal to zero (and $x\geq 0\in \mathbb R$), this means that $x=0$.

This means that any tuple $(x,y,z)$ which solves $f(x,y,z) = 0$, must have the value of $x = 0$.

Answer 2

It's hard to say without you showing even the general structure of your function, but there are a few simple but general strategies:

1. Bounding by a simpler function, using truncated series expansions or inequalities like AM-GM
2. Proving monotonicity in $x$ directly without finding the derivative
3. Splitting cases based on how the graph looks and 'smoothing' towards simple equality cases