A very general method for solving inequalities repaired

Question

Yesterday, I asked a question about a very general method for solving equations I had found here. As it turned out, there were quite some problems with my method and I got a lot of good feedback.

After thinking about it for a while, I've come up with a new method. It still uses the same principle as the old one; it tries to reduce the amount of independent variables based on the assumption that less variables implies an easier to prove inequality.

First, I'll explain the method, then how it avoids some of the things that went wrong with the previous one. After that, I'll ask my questions and at the bottom is an example of how you could prove an inequality with this method.

Edit: Added a tl; dr. If you like what you read there, you can try the technical explanation.

tl; dr

We want to prove $h(a_1,a_2\ldots a_n)\ge0$ with $n\ge 2$. The idea is to pick some function $u$ and fix $u(a_1,a_2)$ and $a_3$ and $a_4, a_5\ldots a_n$ and so on. Then we can only freely change one variable; $a_1$ or $a_2$ since as soon as we change one, we have to change the other in order to make $u(a_1,a_2)$ constant. Now, find the minimum of the $h(a_1,a_2\ldots a_n)$ when $u(a_1,a_2)$ and $a_3,a_4\ldots a_n$ are constant and use that to remove a variable.

A more technical explanation

We want to prove that a certain inequality, $f(A)\ge0$, always holds, with: $$f:D_1\times D_2\ldots\times D_n\to\Bbb{R}$$ And with $D_1,D_2\ldots,D_n\subseteq\Bbb{R}$ and $n\ge 2$.

First, pick two functions $u:D_1\times D_2\to \Bbb{R}$ surjective and $v:D_1\times \Bbb{R} \to D_2$ with: $$\forall a\in D_1\forall b\in D_2:v(a,u(a,b))=b$$ We'll now define an equivalence relation on $\Bbb{D}=D_1\times D_2\ldots\times D_n$. For all $A=(a_1,a_2\ldots a_n),B=(b_1,b_2\ldots b_n)\in \Bbb{D}$, we have $A\sim B$ if and only if: $$u(a_1,a_2)=u(b_1,b_2)$$ $$a_3=b_3,a_4=b_4\ldots a_n=b_n$$ Let $U([A])$ be the value of $u$ associated with $[A]$ and $[A]_3,[A]_4\ldots [A]_n$ the value of $a_3,a_4\ldots a_n$ associated with $[A]$. Now, define $j_{[A]}:D_1\to\Bbb{R}$: $$\forall x\in D_1:j_{[A]}(x)=h(x,v(x,U([A])),[A]_3,[A]_4\ldots [A]_n)$$ Define a function: \begin{align*} E:\Bbb{R}\times D_3\times\ldots\times D_n&\to D_1\\ (U([A]),a_3,a_4\ldots a_n)&\to x:j_{[A]}(x)=\inf\{j_{[A]}(x):x\in D_1\} \end{align*}

Now, assume the inequality $h(A)\ge 0$ doesn't always hold. Let $C=(c_1,c_2\ldots,a_n)\in\Bbb{D}$ be a counterexample. Per definition: $$h(C)\ge h(E(u(c_1,c_2),c_3\ldots c_n),v(E(u(c_1,c_2),c_3\ldots c_n),u(c_1,c_2)),c_3,c_4\ldots c_n)$$ So if $C$ is counterexample, then $C'=(u(c_1,c_2),c_3,\ldots c_n)$ is a counterexample to $h_2(A)\ge 0$, with $A\in\Bbb{R}\times D_3\times\ldots\times D_n$ where: $$h_2(A)=h(E(a_1,a_2\ldots a_{n-1}),v(E(a_1,a_2\ldots a_{n-1}),a_1),a_3,a_4\ldots a_{n-1})$$ Similairly, if $C'$ is a counterexample to $h_2(A)\ge 0$, there must exist at least one combination $c_1,c_2$ such that $c'_1=u(c_1,c_2)$, since $u$ is surjective, so $C=(c_1,c_2,c_3\ldots c_n)$ is a counterexample to $h(A)\ge 0.$ We can conclude that: $$(\forall A\in D_1\times D_2\times\ldots \times D_n:h(A)\ge 0)\Longleftrightarrow(\forall A\in \Bbb{R}\times D_3\times\ldots\times D_n:h_2(A)\ge 0)$$ So we've removed one variable. By induction one can show that eventually, you'll reach an inequality of just one variable and that is what you want.

Some things to keep in mind while using this method

1. If $\inf\{j_{[A]}(x):x\in D_1\}=-\infty$, the inequality clearly isn't true for all inputs and you can stop right there.
2. Finding $\inf\{j_{[A]}(x):x\in D_1\}$ won't always be easy. That's why you can choose $u$ yourself; it might make things easier for you.

How I avoided the problems I had earlier

1. First, we had a whole system of equations you had to solve. Now it's all about finding lots of infimums (do you write it like that?). This means you can go in little steps; if it's too hard to use $a_1$ and $a_2$, change the order of the parameters.
2. I don't assume the infimum is easy to find. I knowingly kept the part of finding $\inf\{j_{[A]}(x):x\in D_1\}$ a bit vague, since I don't really know a general method for finding it (Of course, things like the derivative might still be useful).

Questions

1. First of all, does this method work? (Not about how fast it is, but simply whether is theoretically works)
2. If it works, when would you use it?
3. Is there a smart way to choose $u$, such that $\inf\{j_{[A]}(x):x\in D_1\}$ is easier to find?
4. Can you make some changes to this method, so that it gives you a bijection between counterexamples of $h(A)\ge 0$ and $h_2(A)\ge 0$, where $h_2(A)\ge 0$ is some new inequality with less independent variables?

Example

Prove that for all $a,b,c\in\Bbb{R}$, we have: $$1+\frac13(a+b+c)^2\ge ab+bc+ca+a-c$$ So we get $D_1=D_2=D_3=\Bbb{R}$ and $n=3$. We have: $$h(a,b,c)=1+\frac13(a+b+c)^2-ab-bc-ca-a+c$$ Choose $u(a,b)=a+b$ and $v(a,b)=b-a$. $u$ is surjective and: $$\forall a,b\in\Bbb{R}:v(a,u(a,b))=(a+b)-a=b$$ Now: \begin{align*} j_{[A]}(x)&=h(x,v(x,U([A])),a_3)=h(x,U([A])-x,a_3)\\ &=1 + \frac13(U([A])+a_3)^2 - x(U([A])-x) - (U([A])-x)a_3 - xa_3 - x + a_3\\ &= 1 + \frac13(U([A])+a_3)^2 - x\cdot U([A]) + x^2 - U([A])\cdot a_3 -x + a_3 \end{align*} Now, $1+\frac13(U([A])+a_3)^2-U{[A]}\cdot a_3+a_3$ is constant within $[A]$, so finding $\inf\{j_{[A]}(x):x\in D_1\}$ is really about finding: $$\inf\{x^2-x\cdot U{[A]}-x:x\in D_1\}$$ And that just a simple parabola with its minimum at $\frac12(U([A])+1)$ and that minimum being $0$. We conclude that the value of $x$ such that $j_{[A]}(x)=\inf\{j_{[A]}(x):x\in D_1\}$ is $\frac12(U([A])+1)$, so we define: \begin{align*} E:\Bbb{R}\times D_3&\to D_1\\ (U([A]),a_3)&\to \frac12(U([A])+1) \end{align*} So $h_2(A)$ becomes: \begin{align*} h_2(u,a_3)&=h(E(u,a_3),v(E(u,a_3),u),a_3) = h(\frac{u+1}2,\frac{u-1}2,a_3)\\ &= 1+\frac13(u+a_3)^2 - \frac{u+1}2\cdot\frac{u-1}2 - \frac{u-1}2\cdot a_3 - a_3\cdot\frac{u+1}2 -\frac{u+1}2 + a_3\\ &=1+\frac13(u+a_3)^2 - \frac14(u^2-1+2ua_3-2a_3+2ua_3+2a_3+2u+2-4a_3)\\ &= 1+\frac13(u+a)^2 - \frac14(u^2 + 4ua_3+2u-4a_3+1) \end{align*} So $h_2(u,a_3)\ge0$ becomes: \begin{align*} 1+\frac13(u+a_3)^2 &\ge \frac14(u^2 + 4ua_3+2u-4a_3+1)\\ 12+4(u+a)^2 &\ge 3u^2 + 12ua_3 + 6u - 12a_3 + 3 \end{align*} After choosing some more logical names for our variables, we conclude that, in order to prove the original inequality, we just have to prove that for all $a,b\in\Bbb{R}$: $$12+4(a+b)^2 \ge 3a^2 + 12ab + 6a - 12b + 3 $$ which is easy. Or at least easier than the original equation. If it still looks too hard, we can apply the method again to get an inequality with just one variable.