TITLE: A very general method for proving inequalities. Too good to be true?
QUESTION [62 upvotes]: Update I 'repaired' this method, but it changed a lot and I have some different questions, so I posted it separately here.
As training for the olympiad, I have to solve a lot of inequalities. Recently, I found a very general method to solve inequalities. I think it's too good to be true, but if it is true, I think an implementation of the algorithm in a computer program could be used to very quickly prove all kinds of inequalities. 
First, I'll explain the method. Then I'll ask my questions. At the very bottom, I proved a simple inequality using my method as sort of a demonstration.
The method / algorithm
Say we have to prove a certain inequality that looks like:
$$f(a,b,c\ldots)\ge g(a,b,c\ldots)$$
for all $a,b,c\ldots\in \Bbb{R}$. Here $a,b,c\ldots$ are simply de independent variables. There might be just one, there might be $21$. It doesn't matter.
First, we rearrange:
$$f(a,b,c\ldots)-g(a,b,c\ldots)\ge 0$$
Now, in order to prove our inequality, we have to prove that the LHS has a minimum value and that that minimum value is greater than or equal to $0$. Say that the inequality is true. Let $a_0,b_0,c_0\ldots$ be the values of $a,b,c\ldots$ for which the LHS is minimal. Now, we must have:
$$\frac{d\,(f-g)}{d\, a}(a_0,b_0,c_0\ldots)=0$$
because otherwise, we could increase or decrease $a$ and get a lower value for the LHS, contradicting our definition of $a_0$. In fact, we have:
\begin{align*}
\frac{d\,(f-g)}{d\,a}(a_0,b_0,c_0\ldots) &= 0\\
\frac{d\,(f-g)}{d\,b}(a_0,b_0,c_0\ldots) &= 0\\
\frac{d\,(f-g)}{d\,c}(a_0,b_0,c_0\ldots) &= 0\\
&\vdots
\end{align*}
So we get as many equations as variables. Now we have a nice system of equations to work with. There are three options.

The system has no solutions. This means a minimum does not exist and the inequality doesn't hold for all values of $a,b,c\ldots$.
The system has exactly one solution. If we plug this in and also some other random values for $a,b,v\ldots$, we can see whether it's a maximum or a minimum. If it's a maximum, there are no minima, because the system only had one solution. If it's a minimum, simply plug in the values to check whether the inequality holds. If it does, it holds for all $a,b,c\ldots\in\Bbb{R}$.
The system has infinitely many solutions. In this case, we can express these solutions using less variables than we had in the original inequality. Plugging in those solutions gives a new inequality with less variables, so if we repeat the proces, we eventually get a system of the first or second type or an inequality with just one variable and those are generally pretty easy to solve.

Questions
I have two questions.

Does the method theoretically work? I think so, but there could be something I missed. As I said before, it looks too good to be true.
If it does work, why don't people use it that often? I understand that proofs using things like AM-GM-HM are shorter, but this method is guaranteed to work. It either gives a full proof or a concrete counterexample.

Example
Prove that for all $a,b,c\in\mathbb{R}$, we have:
$$ab+bc+ca+a-c\le 1+\frac13(a+b+c)^2$$
Proof: First, rearrange:
$$1+\frac13(a+b+c)^2-ab-bc-ca-a+c\ge 0$$
Now, take the derivatives with respect to $a$, $b$ and $c$ and set them equal to $0$:
\begin{align*}
\frac23(a+b+c)-b-c-1&=0\\
\frac23(a+b+c) -a-c&= 0\\
\frac23(a+b+c) - b - c + 1 &= 0
\end{align*}
Some algebra to give us a nicer system:
\begin{align*}
b+c+3 &= 2a\\
a+c &= 2b\\
b+c-3 &= 2c 
\end{align*}
So $c = 2b-a$. Plugging it into the first equation gives:
\begin{align*}
b + 2b - a + 3&= 2a\\
3b + 3 &= 3a\\
b &= a-1
\end{align*}
plugging that into the third equation gives:
\begin{align*}
a-1+c-3&=2c\\
a-2 &= c
\end{align*}
and that's the solution to our system; $b=a-1$, $c=a-2$. Plugging this into the original inequality gives:
\begin{align*}
ab+bc+ca+a-c &\le 1+\frac13(a+b+c)^2\\
a(a-1) + (a-1)(a-2) + (a-2)a + a - (a-2) &\le 1+ \frac13(3a-3)^2\\
a^2 - a + a^2 - 3a + 2 + a^2-2a + 2 &\le 1+\frac13(9a^2-18a+9)\\
3a^2 - 6a + 4 &\le 1 + 3a^2 - 6a + 3\\
3a^2 -6a + 4 &\le 3a^2 -6a + 4
\end{align*}
Which always holds and therefore the original inequality always holds. Q.E.D.

REPLY [16 votes]: There are a few issues with this method.

First, it's unclear whether you are trying to solve an inequality or prove an inequality.  They're not the same thing:  Proving an inequality means showing that it's true for all values of the variables (as for example in the inequality $(x+y)^2 \ge 4xy$), whereas solving an inequality means finding the values of the variables for which the inequality is true (as for example in the inequality $x^2 - 1 < 2x$, which is true for $x \in (1-\sqrt{2},1+\sqrt{2})$.

Of course it could be argued that part of solving an inequality always involves a proof, but still it is important to know whether you are trying to show that a given inequality holds for all values of the variable(s), or whether it holds for some values of the variable(s).

You write:


Now, in order to prove our inequality, we have to prove that the LHS has a minimum value and that that minimum value is greater than or equal to 0

This is not true.  For a simple counterexample, consider the inequality $e^{-x}>-2$, which is definitely true for all (real) $x$.  But the function $f(x)=e^{-x}+2$ has no minimum value.  It has a lower bound, even a greatest lower bound (namely $f(x) \ge 2$) but it never actually attains that greatest lower bound.

Even if a function does attain a minimum value, there is no guarantee that the derivative there is $0$.  For example, the function $f(x)=x^{2/3}$ is defined on all of $\mathbb{R}$ and attains its minimum value at $x=0$, but the derivative there is not $0$ because the function isn't differentiable at $x=0$.  In general, remember that critical values of a function occur where the derivative is zero or undefined.
If you are dealing with a function that is defined on some restricted domain (rather than on the whole real line), a minimum value might be attained at an endpoint, rather than at a critical value.  For example if we consider the function $f(x)=\tan(x)$ defined on $[0, \pi/2)$ (which would be the reasonable domain if $x$ represents an angle in a right triangle) then the minimum occurs at $x=0$.

Having said all of this, it is true that if you can find the global minimum of a function, and show that minimum value is positive, that would establish the inequality you are trying to prove.  But that's a pretty big "if".