Find $g(x,y,z);p(x,y,z)\ge 0$ so that $f(x,y,z):=x\cdot g(x,y,z)+p(x,y,z)$ We have the following fact: (I don't remember where I read it, but there is.)

If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$
where $g(x)$ and $h(x)$ are SOS.

So I wanna know how to find $f(x,y,z)$ as $$f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z),$$
where $g(x,y,z),p(x,y,z)$ are all non-negative polynomials.
That also means that my problem is:

Given $f(x,y,z),$ which is a non-negative polynomial where $x,y,z\ge 0.$ Find $g(x,y,z)$ and $p(x,y,z)\ge 0$ (for same condition) such as $f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z).$

There would be something like @Haidangel's question for make

$$H:= m\left ( a, b, c \right )+ kn\left ( a, b, c \right )ca;$$
and $$H:=m\!\left(a,b,c\right)\!-n\!\left(a,b,c\right)\!\left(a-b\right)\!\left(b-c\right)$$
where $m(a,b,c),n(a,b,c)$ are non-negative.

But here, I wanna make it as a sum of polynomials.
I tried $$f(x,y,z)=f(0,y,z)+\left[f(x,y,z)-f(0,y,z)\right]\equiv f(0,y,z)+x\cdot h(x,y,z),$$
But it's not the form I like since $h(x,y,z)$ is not certainly as SOS.
Also, AoPS/@dragonheart6 had told me to use The Unknown coefficients but I don't know how to do it here because there are so many coefficients for the higher degree problems.
Finally, my last purpose is to make $f(x,y,z)$ as SOS where the condition of variables is $x,y,z$ are all non-negative.

$\textbf{Note.}$ We know that if replace $x,y,z$ by $a^2,b^2,c^2$ then we have to prove the inequality for real numbers, but now, the degree is very high.

I wish that someone could have a solution and an example about this.
PS. SOS means Sum of Squares.
 A: 
We have the following fact: (I don't remember where I read it, but there is.)


If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$
where $g(x)$ and $h(x)$ are SOS.

This result can be found in the book by Pólya and Szegö (1st ed, p. 78):

For higher dimensions it is not so easy. You are trying to use Putinar's Positivstellensatz, meaning you are looking for SOS polynomials $g_1$, $g_2$, $g_3$ and $p$ such that:
$$f(x,y,z)=x\cdot g_1(x,y,z)+y\cdot g_2(x,y,z)+z\cdot g_3(x,y,z)+p(x,y,z).$$
The conditions for that theorem are not satisfied for nonnegativity over $\mathbb{R}^n_+$, so not every nonnegative $f$ will admit such a representation with $g_i$ and $p$.
The Krivine–Stengle Positivstellensatz states that every polynomial $f$ that satisfies $f(x)\geq 0$ for all $x\in\mathbb{R}^3_+$ can be decomposed using SOS polynomials $g_1,\ldots,g_6$ as:
$$f(x)\left(x_1\cdot g_4(x)+x_2\cdot g_5(x)+x_3\cdot g_6(x)\right) = f(x)^{2k} + x_1\cdot g_1(x)+x_2\cdot g_2(x)+x_3\cdot g_3(x)$$
for some $k\in\mathbb{N}$. The classic method to find $g_1,\ldots,g_6$ is using semidefinite optimization (which was already mentioned in the comments). YALMIP is another excellent tool for that. This method requires you to have an upper bound on the degree of the polynomials, which you do not have. You also do not know $k$. What you could do is vary $k$ and gradually increase the upper bound and hope you get 'lucky'. I am not aware of any other general method.
A nice background read is Section 3 (3.3 and 3.6 in particular) of Sums of squares, moment matrices and optimization over polynomials by M. Laurent.
