Let $X$ be a non-negative random variable and $\theta \in [0,1]$. Show that:
$$\mathbb P\left(X \geq \theta \, \mathbb E X\right ) \geq (1-\theta)^2\frac{\mathbb E [X]^2}{\mathbb E[ X^2] - \theta(2-\theta)\mathbb E[X]^2}$$
What I've tried
This is very similar to the Paley-Zygmund inequality (in which case the denominator is missing the second term). I don't think the proof for PZ is adaptable here (unless we can apply something stronger than Cauchy-Schwarz):
$$\mathbb E X = \mathbb E X \mathbb 1(X < \theta \mathbb E X) + \mathbb E X \mathbb 1(X\geq \theta \mathbb E X) \leq \theta \mathbb E X + \sqrt {\mathbb E X \mathbb P(X \geq\theta \, \mathbb EX)}$$
Where the inequality is from cauchy-schwarz. Simplifying gives the PZ inequality. If I can get a tighter bound on the second term then I should be able to show the inequality of interest.
Any tips would be appreciated.
Note that your denominator is exactly $\mathbb{E}[(X - \theta \mathbb{E}[X])^2]$. So we have the following:
$$ \begin{aligned} (1 - \theta) \mathbb{E}[X] &= \mathbb{E}[X - \theta E[X]] \leq \mathbb{E}\big[(X - \theta \mathbb{E}[X]) 1\{X - \theta \mathbb{E}[X] \geq 0\}\big] \\ &\overset{\text{(C.S.)}}{\leq} \sqrt{\mathbb{P}(X \geq \theta \mathbb{E}[X])} \sqrt{\mathbb{E}[(X - \theta \mathbb{E}[X])^2]} \end{aligned} $$
Rearranging will give you the result.