The main idea is correct. Actually, the derivative is not messy with careful factoring.
Suppose $a$ is fixed. Define
$$f(x)=ax^{2a}-2ax^a-x^{a+1}+(a+1)x$$
Since you have found that $f(1)=0$ and $f(0)=0$, we may think we can do some thing with its monoticity! If we can prove $f(x)$ increases at first and decreases to $0$ when $x=1$,
$f(x)\ge 0$ will hold.
Thus its derivative can help us a lot:
\begin{align}
f'(x)&=2a^2x^{2a-1}-2a^2x^{a-1}-(a+1)x^a+a+1\\
&=2a^2x^{a-1}(x^a-1)-(a+1)(x^a-1)\\
&=(2a^2x^{a-1}-a-1)(x^a-1)
\end{align}
Define two new function $h(x)=2a^2x^{a-1}-a-1$ and $g(x)=x^a-1$, and then see how their value goes on $[0,1]$
Of course, $g(x)\le 0$ on $[0,1]$. While as to $h(x)$, it is an increasing fuction on $[0,1]$ and $h(0)=-a-1<0$, $h(1)=2a^2-a-1=(2a+1)(a-1)\ge0$. In this way, we should be able to find an $0<x_0<1$ such that $h(x_0)=0$. Therefore $h(x)\ge0$ on $[0,x_0)$ and $h(x)\le0$ on $(x_0,1]$.
Since $f'(x)=g(x)h(x)$, we can get that $f'(x)\ge0$ on $[0,x_0)$ and $f'(x)\le0$ on $(x_0,1]$. Thus, $f(x)$ will increase on $[0,x_0)$ and decrease on $(x_0,1]$, which fits the assumption we made at the beginning!
Factoring can help a lot in this kind of questions.