Alternate proof of the integral: $\int_0^1 x^x(1-x)^{2x}\,dx\neq3/8$ I am looking into the integral: $$I=\int_0^1 x^x(1-x)^{2x}\,dx\neq\frac{3}{8}$$
How might you prove this to be true? What's tough is that the integral $$3/8\lt I<0.37503$$ numerically. I managed to prove this by means of Riemann sums, classically, but ideas like Taylor expansion are extremely difficult in this case...
 A: This is not a solution but approximation. First, note that $\frac{3}{8}=0.3750000000$. The function $f(x)=x^x(1-x)^{2x}$ is convex on $(0,c_1)\cup(c_2,1)$ and concave on $(c_1,c_2)$, where $c_1\approx  0.2718247$ and $c_2\approx 0.5243816$.
So that applying the hermite-hadamard inequality 
\begin{eqnarray}
f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b -
a}}\int\limits_a^b {f\left( x \right)dx}  \le \frac{{f\left( a
\right) + f\left( b \right)}}{2},
\end{eqnarray}
which hold for all convex functions $f$ defined on a real interval
$[a,b]$. The inequality is reversed if $f$ is concave. The inequality is sharp in both sides.
In our case $[a,b]=[0,1]$.
Therefore, On $(0,c_1)$, we have
\begin{eqnarray}
 f\left( {\frac{{0 + c_1}}{2}} \right) \le \frac{1}{{c_1 -
0}}\int\limits_0^{c_1} {f\left( x \right)dx}  \le \frac{{f\left( 0
\right) + f\left( c_1 \right)}}{2}
\end{eqnarray}
and we write
\begin{eqnarray}
0.1991787227\le  \int\limits_0^{c_1} {f\left( x \right)dx}  \le 0.2149827495\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)
\end{eqnarray}
On $(c_2,1)$, we have
\begin{eqnarray}
0.04452261402\le \int\limits_{c_2}^{1} {f\left( x \right)dx} \le 0.07962031056\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{eqnarray}
On $(c_1,c_2)$, $f$ is concave and thus the Hermite-Hadamard inequality is reversed, then we have
\begin{eqnarray}
0.1155270131\le \int\limits_{c_1}^{c_2} {f\left( x \right)dx} \le 0.1164615612\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)
\end{eqnarray}
Adding the inequalities (1)-(3), we get
\begin{eqnarray}
0.3581809649\le \int\limits_{0}^{1} {f\left( x \right)dx} \le 0.4110646213 
\end{eqnarray}
Due to sharpness of H.-H. inequality, this is the best possible analytic  approximation even that the approximation of $c_1,c_2$ is almost accurate.
Using Maple I get the numerical solution 0.3750261533 which is not $\frac{3}{8}$.
