Prove: When $a=0$, the equation becomes $2 b x=b$. Then, regardless of the value of $b$, there is always a root $x=\frac{1}{2} \in(0,1)$.
When $a \neq 0$, let $f(x)=3 a x^{2}+2 b x-(a+b)$.
Below, find $x_{1}<x_{2}$, such that
$$f\left(x_{1}\right) \cdot f\left(x_{2}\right) \leqslant 0 \text{. . }$$
$$\begin{array}{l} \text { Two } f\left(x_{1}\right)=3 a x_{1}^{2}+2 b x_{1}-(a+b), \\ f\left(x_{2}\right)=3 a x_{2}^{2}+2 b x_{2}-(a+b) . \\ \text { Let } f\left(x_{1}\right)=-f\left(x_{2}\right) \text{. Then } \\ 3 a x_{1}^{2}+2 b x_{1}-(a+b) \\ =-3 a x_{2}^{2}-2 b x_{2}+(a+b), \end{array}$$

i.e., $\left(3 x_{1}^{2}+3 x_{2}^{2}-2\right) u+\left(2 x_{1}+2 x_{2}-2\right) b=0$.
The above equation holds for different values of $a$ and $b$, then
$$\left\{\begin{array}{l} 3 x_{1}^{2}+3 x_{2}^{2}-2=0 \\ 2 x_{1}+2 x_{2}-2=0 \end{array}\right.$$

Solving, we get $x_{1}=\frac{3-\sqrt{3}}{6}, x_{2}=\frac{3+\sqrt{3}}{6}$.
Thus, we have found $x_{1}<x_{2}$, such that
$$\begin{array}{l} f\left(x_{1}\right) f\left(x_{2}\right)=f\left(x_{1}\right)\left[-f\left(x_{1}\right)\right] \\ =-f^{2}\left(x_{1}\right) \leqslant 0 \end{array}$$

is satisfied.
At this time, by $0<\frac{3-\sqrt{3}}{6}<\frac{3-\sqrt{3}}{6}<1$, if
$$f\left(x_{1}\right) f\left(x_{2}\right)=0 \text{, }$$

then it is proven that $x_{1}$ and $x_{2}$ are both roots of the original equation, and the proposition is established.
If $f\left(x_{1}\right) f\left(x_{2}\right)<0$, by the graph of a linear equation, it is known that $f(x)=3 a x^{2}+2 b x-(a+b)$ intersects the $x$-axis at one point in $\left(x_{1}, x_{2}\right)$. This indicates that the original equation has one real root in $\left(x_{1}, x_{2}\right)$, hence the original equation has one real root in $(0,1)$, and the proposition is established.
In summary, the proposition is proven.
(Yang Yongliang, Yangtze River No.5 Middle School, Yueyang City, Hunan Province, 414411)