Think of the expression given as distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in $\Bbb{R^3}$. So let us figure out what are $y_i$ and $z_i$. For example, we can take $y_1=2-x_1$ and $y_2=x_2$ and so on...
Consider the lines $L_1$ and $L_2$ given by the vector equations
$$L_1: \mathbf{r}_1=\begin{bmatrix}t\\2-t\\2t\end{bmatrix}=\underbrace{\begin{bmatrix}0\\2\\0\end{bmatrix}}_{\color{red}{\mathbf{a}}}+t\underbrace{\begin{bmatrix}1\\-1\\2\end{bmatrix}}_{\color{red}{t\mathbf{b}}} \quad \text{ and } \quad L_2: \mathbf{r}_2=s\underbrace{\begin{bmatrix}1\\1\\3\end{bmatrix}}_{\color{blue}{s\mathbf{d}}}$$
Then the shortest distance between these two lines is what you are looking for. Perhaps you can take it from here and since the lines are not parallel you can use shortest distance between two skew lines
Some helpful details:
The unit normal vector $\vec{\mathbf{n}}=\begin{bmatrix}-\sqrt{\frac{5}{6}}\\-\frac{1}{\sqrt{30}}\\\sqrt{\frac{2}{15}}\end{bmatrix}$ and if my calculations are correct then the minimum value is $\color{magenta}{\sqrt{\frac{2}{15}}}$.
Some thoughts.
Fact 1. $1 - xy\mathrm{e}^{y - 2x} > 0$ on
$(\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$.
Fact 2. For each fixed $x > \frac{1}{\sqrt{2}}$, the equation $
\ln y - x \mathrm{e}^{y-2x} = \ln 2 + \ln x - x$ has exactly one real solution $y \in (0, \sqrt{2})$.
$\phantom{2}$
Now, let
$$F(x, y) := \ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x).$$
Clearly, $F(x, y)$ is continuously differentiable on $(\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$. By Fact 1, we have
$$\frac{\partial F}{\partial y} = \frac{1}{y}\left(1 - xy\mathrm{e}^{y - 2x}\right) > 0.$$
By the Implicit Function Theorem, using Fact 2, the equation $F(x, y) = 0$ implicitly determines $y$ as a differentiable function of $x$, given that $x\in (\frac{1}{\sqrt{2}}, \infty)$.
Taking derivative with respect to $x$ on
$\ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x) = 0$, we have
$$\frac{1}{y} \cdot y'(x)
- \mathrm{e}^{y-2x} - x \mathrm{e}^{y-2x}(y'(x) - 2)
- \frac{1}{x} + 1 = 0$$
or
$$\frac{1}{y}\left(1 - xy\mathrm{e}^{y - 2x}\right)y'(x) = \frac{1}{x} + (1 - 2x)\mathrm{e}^{y - 2x} - 1. \tag{1}$$
From (1), using Fact 1, it suffices to prove that
$$\frac{1}{x} + (1 - 2x)\mathrm{e}^{y - 2x} - 1 < 0,$$
or
$$\mathrm{e}^{y - 2x} > \frac{1/x - 1}{2x - 1}\tag{2}$$
given that $(x, y) \in (\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$ with $\ln y - x \mathrm{e}^{y-2x} = (\ln 2 + \ln x - x)$.
Clearly, we only need to prove the case that $\frac{1}{\sqrt{2}} < x < 1$. (2) is written as
$$y > 2x + \ln \frac{1/x - 1}{2x - 1}. \tag{3}$$
Clearly, we only need to prove the case that $2x + \ln \frac{1/x - 1}{2x - 1} > 0$. Note that
$2x + \ln \frac{1/x - 1}{2x - 1} < \sqrt{2}$
for all $\frac{1}{\sqrt{2}} < x < 1$ (easy). By Fact 2, we have $y < \sqrt{2}$.
Let
$f(y) := \ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x)$. By Fact 1, we have $f'(y) > 0$ on $(0, \sqrt{2})$, given that $x \in (\frac{1}{\sqrt{2}}, \infty)$.
Thus, we have
$$f(y) > f\left(2x + \ln \frac{1/x - 1}{2x - 1}\right)
\iff y > 2x + \ln \frac{1/x - 1}{2x - 1}$$
given that $2x + \ln \frac{1/x - 1}{2x - 1}\in (0, \sqrt{2})$, and $y \in (0, \sqrt{2})$. Using $f(y) = 0$, it suffices to prove that
$$0 > f\left(2x + \ln \frac{1/x - 1}{2x - 1}\right),$$
or
$$\ln \left(2x + \ln \frac{1/x - 1}{2x - 1}\right) - x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 0 \tag{4}$$
for all $x \in (\frac{1}{\sqrt{2}}, 1)$ with $2x + \ln \frac{1/x - 1}{2x - 1} > 0$.
Note that $- x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 1 - \ln 2$ for all $\frac{1}{\sqrt{2}} < x < 1$.
We only need to prove the case that
$\ln \left(2x + \ln \frac{1/x - 1}{2x - 1}\right) + (1 - \ln 2) \ge 0$,
i.e. $2x + \ln \frac{1/x - 1}{2x - 1} \ge \frac{2}{\mathrm{e}}$
which implies $x < 4/5$. Thus, it suffices to prove (4)
for $x \in (\frac{1}{\sqrt{2}}, \frac45)$.
Using $\ln (1 + u) \ge \frac{u}{2}\cdot \frac{2 + u}{1 + u}$ for all $-1 < u \le 0$, using $\frac{1/x - 1}{2x - 1} \in (0, 1)$, we have
$$\ln \frac{1/x - 1}{2x - 1} \ge
- \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)}.$$
It suffices to prove that
$$g(x) := \ln \left(2x - \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)}\right) - x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 0.$$
(Note: $2x - \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)} > 0$ for all $x \in (\frac{1}{\sqrt{2}}, \frac45)$.)
We have
$$g'(x) = \frac{(2x^2 - 1)(24x^6 - 80x^5 + 116x^4 - 102x^3 + 56x^2 - 17x + 2)}{x(1 - x)(2x - 1)^2(-12x^4 + 16x^3 - 4x^2 - 2x + 1)} < 0.$$
Also, we have $g(\frac{1}{\sqrt{2}}) = 0$. Thus, $g(x) < 0$ for all $x \in (\frac{1}{\sqrt{2}}, \frac45)$.
Best Answer
Applying AM-GM, you have to consider if the equality can hold. If not, you only get a strictly lower bound rather than the minimum.
We may use AM-GM in this way: $$x_1^2 + y_1^2 + x_2^2 + y_2^2 + (- x_1x_2) + (- x_1x_2) + (- y_1 y_2) + (- y_1y_2) \ge 8\sqrt[8]{x_1^4 x_2^4 y_1^4 y_2^4} = 8$$ with equality if $x_1 = y_1 = 1, \ x_2 = y_2 = -1$.