By symmetry, it is enough to show that \begin{equation} f(x):=x^{2-2 x}+(1-x)^{2 x}\le1 \end{equation} for \begin{equation} x\in(0,\tfrac12); \end{equation} the latter condition will be assumed by default. It is easy to see that \begin{equation} r_0(x):=1 - x - x \ln x>0. \end{equation} So, $f'(x)$ equals \begin{equation} f_1(x):=\frac{f'(x)}{x^{1 - 2 x} r_0(x)} \end{equation} in sign. Moreover, \begin{equation} f_2(x):=f'_1(x)\frac{r_0(x)^2}{2 \big((1 - x) x\big)^{2 x}} \end{equation} is a rational function of $x,\ln x,\ln(1-x)$, and $f_2(x)$ equals $f'_1(x)$ in sign.
This almost completes the solution, since the sign pattern of such a rational function can be determined completely algorithmically, and then one can go back and determine the sign patterns of $f_1$ and $f$$f-1$, in this order. So, the only remaining problem is to find a better algorithm.
Differentiating $f_2(x)$, we kill the terms containing $\ln^2 x\ln(1-x)$ and $\ln x\ln^2(1-x)$, so that
\begin{equation}
f_3(x):=f'_2(x)/c_3(x)
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x),\ln x\ln(1-x)$ (with coefficients in $\mathbb Z(x)$), where
\begin{equation}
c_3(x):=\frac{4 \left(2 x^3-3 x^2-x+1\right)}{(1-x)^2 x^2}>0.
\end{equation}
Next,
\begin{equation}
f_4(x):=f'_3(x)\,\frac{(1 - x)^3 x^3 c_3(x)^2}{4 - x + x^2}
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x)$
and
\begin{equation}
f_5(x):=f'_4(x)\,\frac{x^2 (4 - x + x^2)^2}{2c_3(x)(1-x)c_5(x)}
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln(1-x)$, where
$c_5(x):=12 + 4 x + 47 x^3 - 99 x^4 + 10 x^5 - 6 x^6>0$.
Further,
\begin{equation}
f_6(x):=f'_5(x)\, \frac{c_5(x)^2 (1 - x)^4}{x (4 - x + x^2)}
\end{equation}
is a polynomial of degree 1 in $\ln x$, and hence so is
$f'_6(x)$. Therefore, the value of $f'_6(x)$ is between those of $f_{73}(x)$ and $f_{74}(x)$, where $f_{7k}(x)$ is obtained from $f'_6(x)$ by replacing there $\ln x$ with the Taylor polynomial of degree $k$ for $\ln x$ at the point $1/2$.
Since $f_{73}(x)$ and $f_{74}(x)$ are rational expressions (in $\mathbb R(x)$), it is easy to see that $f_{73}<0$, $f_{74}<0$, and hence $f'_6<0$, so that $f_6$ decreases (on $(0,1/2)$).
Moreover, $f_6(0+)>0>f_6(1/2)$. So, $f_6$ is $+-$; that is, it changes in sign only once on $(0,1/2)$, from $+$ to $-$.
So, $f_5$ is up-down; that is, it increases on $(0,c)$ and decreases on $(c,1/2)$, for some $c\in(0,1/2)$. Also, $f_5(0+)=0=f_5(1/2)$. So, $f_5>0$, $f_4$ increases, to $f_4(1/2)=0$. So, $f_4<0$, $f_3$ decreases, from $f_3(0+)=0$. So, $f_3<0$, $f_2$ decreases, from $f_2(0+)=\infty$ to $f_2(1/2)=-1<0$. So, $f_2$ is $+-$, $f_1$ is up-down, with $f_1(0+)=-2<0$ toand $f_1(1/2)=0$. So, $f_1$ is $-+$, $f$ is down-up, with $f(0+)=f(1/2)=1$. So, $f<1$ on $(0,1/2)$. QED