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 $f-1$, 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 $f_1(0+)=-2<0$ and $f_1(1/2)=0$.
So, $f_1$ is $-+$, $f$ is down-up, with $f(0+)=f(1/2)=1$.
So, $f<1$ on $(0,1/2)$. QED

Always ask a question in your titleand 2) try to make that question as close as possible an approximation to the full question you're asking. Unfortunately everyone uses the title more as an email subject line. $\endgroup$ – Scott Morrison♦ Mar 5 '10 at 18:36