China-TST-A

Denote the given polynomial by $F(x, y, z)$. By contradiction, assume there exist real-coefficient polynomials $f_1(x, y, z)$, $f_2(x, y, z)$, ..., $f_m(x, y, z)$ satisfying
$$F(x, y, z) = \sum_{i=1}^{m} f_i(x, y, z)^2. \qquad (1)$$
First, all $f_i$ must have degree at most 3. If some $f_i$ had degree $d > 3$, then the sum of squares of their degree $d$ homogeneous parts would be 0, implying all these parts vanish, which is a contradiction!
Thus all $\deg(f_i) \le 3$. By replacing each $f_i(x, y, z)$ with its degree 3 homogeneous part, we may assume each $f_i(x, y, z)$ is a homogeneous real-coefficient polynomial of degree 3.
Setting $x = y$ in (1) yields
$$z^4(z-y)^2 = \sum_{i=1}^{m} f_i(y, y, z)^2. \qquad (2)$$
Let $g_i(y, z) = f_i(y, y, z)$. Setting $y = z$ in (2) gives $0 = \sum_{i=1}^{m} g_i(y, y)^2$, so $g_i(y, y) = 0$ for all $i$. Thus $z - y \mid g_i(y, z)$; similarly, setting $z = 0$ shows $z \mid g_i(y, z)$. Therefore we can write $g_i(y, z) = z(z - y)h_i(y, z)$, where $h_i(y, z)$ is linear in $y$ and $z$. Substituting into (2) and canceling $z^2(z - y)^2$ gives
$$z^2 = \sum_{i=1}^{m} h_i(y, z)^2.$$
Setting $z = 0$ shows $h_i(y, z) = \alpha_i z$, so $f_i(y, y, z) = \alpha_i z^2(z - y)$, where $\alpha_i \in \mathbb{R}$. In particular, the coefficients of $y^3$ and $y^2z$ in $f_i(y, y, z)$ are 0, and the sum of coefficients of $z^3$ and $z^2y$ is 0.
Now fix one $f_i$ and write $f_i = \sum_{\substack{u,v,w \ge 0 \\ u+v+w=3}} a_{u,v,w} x^u y^v z^w$. Then:
$$\left\{ \begin{array}{l} a_{3,0,0} + a_{2,1,0} + a_{1,2,0} + a_{0,3,0} = 0; \\ a_{2,0,1} + a_{1,1,1} + a_{0,2,1} = 0; \\ a_{1,0,2} + a_{0,1,2} + a_{0,0,3} = 0. \end{array} \right. \qquad (3)$$
These relations can be visualized in Figure 1, where dashed boxes enclose elements summing to 0.

Figure 1

By symmetry, we also have:
$$\left\{ \begin{array}{l} a_{3,0,0} + a_{2,0,1} + a_{1,0,2} + a_{0,0,3} = 0; \\ a_{0,3,0} + a_{0,2,1} + a_{0,1,2} + a_{0,0,3} = 0; \\ a_{1,2,0} + a_{1,1,1} + a_{1,0,2} = 0; \\ a_{2,1,0} + a_{1,1,1} + a_{0,1,2} = 0. \end{array} \right. \qquad (4)$$
In the figure, this corresponds to sums of elements along edges and diagonals being 0.
From (3), the sum of all $a_{u,v,w}$ is 0. Removing the corner triangles shows $a_{1,1,1} = 0$.
Then (4) implies:
$$a_{2,1,0} + a_{0,1,2} = 0, \quad a_{2,0,1} + a_{0,2,1} = 0, \quad a_{1,2,0} + a_{1,0,2} = 0.$$
Let $a = a_{2,1,0}$, $b = a_{0,2,1}$, $c = a_{1,0,2}$, then $a_{0,1,2} = -a$, $a_{2,0,1} = -b$, $a_{1,2,0} = -c$.
Now, the coefficient of $x^2y^2z^2$ in $f_i(x, y, z)^2$ is:
$$a_{1,1,1}^2 + 2a_{2,1,0} \cdot a_{0,1,2} + 2a_{2,0,1} \cdot a_{0,2,1} + 2a_{1,2,0} \cdot a_{1,0,2} = -2a^2 - 2b^2 - 2c^2 \le 0,$$
while $F(x, y, z)$ has coefficient 0 for $x^2y^2z^2$. From (1), we must have $a = b = c = 0$ for
each $f_i$. Substituting back into (3) and (4) shows $f_i = 0$, contradicting (1). Therefore,
$F(x, y, z)$ cannot be expressed as a finite sum of squares. $\square$

Denote the polynomial by $F(x, y, z)$. By contradiction, assume $F$ can be written as a sum of squares of real-coefficient polynomials. Then there exist $f_i(x, y, z), g_i(x, y, z) \in \mathbb{R}[x, y, z]$ ($i = 1, \dots, m$) with:
$$F = \sum_{i=1}^{m} (x f_i + g_i)^2 = \sum_{i=1}^{m} x^2 f_i^2 + \sum_{i=1}^{m} g_i^2 + 2x \sum_{i=1}^{m} f_i g_i, \quad (5)$$
where each monomial in $f_i(x, y, z)$, $g_i(x, y, z)$ has even degree in $x$. Since $\deg(F) = 6$, as
in Proof 1 we may assume $f_i, g_i$ are homogeneous of degrees 2 and 3 respectively, with
$\deg_x f_i \le 2$, $\deg_x g_i \le 2$.

Setting $(x, y, z) = (0, 1, 1)$ in (5) gives:
$$\sum_{i=1}^{m} g_{i}^{2}(0, 1, 1) = 0,$$
so $g_i(0, 1, 1) = 0$. Thus we can write $g_i = c_i x^2 + (y-z)d_i$ for some $c_i, d_i \in \mathbb{R}[y, z]$.
Setting $y = z = 1$ in (5) and comparing coefficients of $x^2$ gives:
$$\sum_{i=1}^{m} f_{i}^{2}(0, 1, 1) = 0,$$
so $f_i(0, 1, 1) = 0$. Thus $f_i = a_i x^2 + (y-z)b_i$ for some $b_i \in \mathbb{R}[y, z]$ and $a_i \in \mathbb{R}$.
Using the identity:
$$(ax^3 + b)^2 + (cx^3 + d)^2 = \left(\sqrt{a^2 + c^2}x^3 + \frac{ab + cd}{\sqrt{a^2 + c^2}}\right)^2 + \left(\frac{ad - bc}{\sqrt{a^2 + c^2}}\right)^2,$$
we may assume $a_1 = 1$ and $a_i = 0$ for $i \ge 2$. Comparing coefficients of $x^5$ gives:
$$2 \sum_{i=1}^{m} a_i c_i = -y - z,$$
so $c_1 = -\frac{y+z}{2}$.
Comparing coefficients of $x^4$ gives:
$$2(y - z)b_1 + \sum_{i=1}^{m} c_i^2 = yz.$$
Thus:
$$2(y - z)b_1 + \frac{1}{4}(y - z)^2 + \sum_{i=2}^{m} c_i^2 = 0.$$
Setting $y = z$ shows $y - z \mid c_i$ for $i \ge 2$.

Comparing coefficients of $x^2$ gives:
$$(y - z) \sum_{i=1}^{m} b_i^2 + 2 \sum_{i=1}^{m} c_i d_i = 0.$$
Setting $y = z$ shows $y - z \mid d_1$. Let $d_1 = (y - z)(sy + tz)$ for $s, t \in \mathbb{R}$.
Comparing coefficients of $x^3$ gives:
$$d_1 + \sum_{i=1}^{m} b_i c_i = 0.$$
Setting $y = z$ shows $y - z \mid b_1$. Let $b_1 = k(y - z)$ for $k \in \mathbb{R}$.
From $F(1, 1, 0) = 0$, we get $1 + k - \frac{1}{2} + s = 0$, so $s = -\frac{1}{2} - k$. From $F(1, 0, 1) = 0$, we get $1 + k - \frac{1}{2} + t = 0$, so $t = -\frac{1}{2} - k$.
Thus:
$$x f_1 + g_1 = x(x^2 + k(y - z)^2) - \frac{y+z}{2}x^2 - (\frac{1}{2} + k)(y + z)(y - z)^2.$$

Setting $x = 1, y = 1$ in $F - (xf_1 + g_1)^2 \ge 0$ gives:
$$\begin{aligned} 0 &\le z^4(z-1)^2 - \left(\frac{1-z}{2} + k(1-z)^2 - \left(\frac{1}{2} + k\right)(1-z)^2(1+z)\right)^2 \\ &= (z-1)^2 \left[z^4 - \left(\frac{1}{2} + k(1-z) - \left(\frac{1}{2} + k\right)(1-z^2)\right)^2\right]. \end{aligned}$$
The inequality $z^2 - ((\frac{1}{2} + k)z - k)^2 \ge 0$ holds for all real $z$ only when $k = 0$. But then setting $z = 0, x = 7, y = 2$ gives $F(x, y, z) - (xf_1 + g_1)^2 = -225 < 0$, a contradiction.
Therefore, $F(x, y, z)$ cannot be expressed as a finite sum of squares of real-coefficient polynomials. $\square$