Prove that $\mathbb{P}(X > \lambda \mathbb{E}[X]) \geq (1-\lambda)^2\frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}$ I am trying to prove that $\mathbb{P}(X > \lambda \mathbb{E}[X]) \geq (1-\lambda)^2\frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}$ for all non-negative random variables X with $0 \leq \lambda < 1$.
I have tried many different paths, however the most successful one so far involves the Cauchy-Schwarz inequality, which states:
$$\mathbb{E}[X Y] \leq \sqrt{\mathbb{E}[X^2]\mathbb{E}[Y^2]}$$
This implies:
$$\mathbb{E}[Y^2] \geq \frac{\mathbb{E}[X Y]^2}{\mathbb{E}[X^2]}$$
Now I think with a proper choice for Y I might be able to obtain the statement I am trying to prove. I tried for instance $Y=1-\lambda$ which I believe leads to:
$$(1-\lambda)^2 \geq (1-\lambda)^2\frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]}$$
This is the closest I got to the statement, however now I would have to prove that $\mathbb{P}(X > \lambda \mathbb{E}[X]) \geq (1-\lambda)^2$ at which I got stuck.
Am I trying the right things here or am I completely going down a dead path? If so, what could be a sensible choice for $Y$? Thanks in advance!
 A: Cauchy Schwarz inequality is a good idea.


*

*Show using the Cauchy Schwarz inequality that
$$\mathbb{E}(X 1_{\{X \geq \lambda \mathbb{E}X\}}) \leq \sqrt{\mathbb{E}(X^2)} \sqrt{\mathbb{P}(X \geq \lambda \mathbb{E}X)}. \tag{1}$$

*Show that $$\mathbb{E}(X 1_{\{X \geq \lambda \mathbb{E}X\}}) \geq (1-\lambda) \mathbb{E}(X). \tag{2}$$ (Hint: $1_{\{X \geq \lambda \mathbb{E}X\}}) = 1- 1_{\{X < \lambda \mathbb{E}X\}})$.)

*Combine $(1)$ and $(2)$ to prove the claim.


Remark: This inequality is known as Paley-Zygmund inequality.
A: I think there is a little bit clearer proof of the Paley–Zygmund inequality on Wikipedia. The idea is the same, but it is a little bit clearer. The proof goes as follows.
First, 
$$
\operatorname{E}[Z] = \operatorname{E}[ Z \, \mathbf{1}_{\{ Z \le \theta \operatorname{E}[Z] \}}]  + \operatorname{E}[ Z \, \mathbf{1}_{\{ Z > \theta \operatorname{E}[Z] \}} ].
$$
The first addend is at most $\theta\operatorname{E}[Z]$, while the second is at most $\operatorname{E}[Z^2]^{1/2} \operatorname{P}( Z > \theta\operatorname{E}[Z])^{1/2}$ by the Cauchy–Schwarz inequality. The desired inequality then follows.
