Let $f(x) = \ln x + x - ax^2$, where $a > 0$. Suppose $x_1 < x_2$ are two zeros of $f(x)$, prove that$$ f'\left( \frac{x_1 + 2x_2}{3} \right) < 1 - a. $$
My try:
Denote $x_3 = \dfrac{x_1 + 2x_2}{3}$. From $f(x_1) = f(x_2) = 0$, there is$$ \frac{\ln x_1 + x_1}{x_1^2} = \frac{\ln x_2 + x_2}{x_2^2} = a. $$ Also,$$ f'(x) = \frac{1}{x} + 1 - 2ax. $$ Thus\begin{align*} &\mathrel{\phantom{\Longleftrightarrow}} f'(x_3) < 1 - a \Longleftrightarrow 2ax_3^2 - ax_3 - 1 > 0 \Longleftrightarrow ax_3 (2x_3 - 1) > 1\\ &\Longleftrightarrow \left( \frac{1}{3} ax_1 + \frac{2}{3} ax_2 \right) \left( \frac{2}{3} x_1 + \frac{4}{3} x_2 - 1 \right) > 1\\ &\Longleftrightarrow \left( \frac{1}{3} \frac{\ln x_1 + x_1}{x_1} + \frac{2}{3} \frac{\ln x_2 + x_2}{x_2} \right) \left( \frac{2}{3} x_1 + \frac{4}{3} x_2 - 1 \right) > 1\\ &\Longleftrightarrow \left( \frac{\ln x_1}{x_1} + \frac{2\ln x_2}{x_2} + 3 \right) (2x_1 + 4x_2 - 3) > 9. \end{align*}
Then I tried to prove the last inequality for arbitrary $x_1 < x_2$, but the expression has gotten really complicated so far and the logarithm terms make it even worse to handle.
Is there a better approach to this question? Thanks in advance.
Edit: From the comment of @MartinR and some experiment with the graph of $f$, it seems promising to prove that $0 < a < 1$ and $f'\left( \dfrac{x_1 + 2x_2}{3} \right) < 0$. Furthermore, it also seems true that $f'\left( \dfrac{x_1 + x_2}{2} \right) < 0$, which is stronger than the original inequality to be proved in that $f'$ is decreasing. So it needs proving that$$ f'\left( \frac{x_1 + x_2}{2} \right) = \frac{2}{x_1 + x_2} + 1 - a(x_1 + x_2) < 0. $$ Any ideas?