How to prove the following inequality (with a introduced constant $\lambda$)? Suppose that $X$ is a positive random variable, with expectation $\mu$ and variance $\sigma^2$.
Prove, for all $\lambda\in(0,1)$, we always have
$$P(X>\lambda \mu) \geqslant (1-\lambda)^2 \frac{\mu^2}{\mathbb{E}[X^2]}$$
Can someone sort out the proof using Chebyshev's inequality? Thanks very much.
 A: Note that $X$ can be expressed as a sum, since the space can be partitioned, such that $\mu = \mathbb{E}[X] = \mathbb{E}[X \cdot I_{\{X \le \lambda\mu\}}]+\mathbb{E}[X \cdot I_{\{X  > \lambda\mu\}}]$, due to the linearity of the expected value.
The lefthand summand is bounded by $\lambda\mu$, subtracting this, rearranging, and squaring both sides, we equivalently have
$$\mu^2(1-\lambda)^2 \le \big(\mathbb{E}[X\cdot I_{\{X  > \lambda\mu\}}]\big)^2.$$
As the righthand integrand is almost surely positive, we can apply the Cauchy-Schwarz inequality to get 
$$\mu^2(1-\lambda)^2 \le \mathbb{E}[X^2]\cdot\mathbb{E}[I^2_{\{X  > \lambda\mu\}}] = \mathbb{E}[X^2] \cdot \mathbb{P}(X > \lambda\mu),$$
as we note that the indicator function is invariant under the square function and its expected value is the probability of the indexed event.
Lastly, note that due to positivity of $X$, by monotony of the expected value and $\mathbb{P}(X=0)=0$, $\mu$ is positive. Hence, even if $\sigma^2 = \mathbb{E}[X^2]-\mathbb{E}[X]^2 = 0$, still $\mathbb{E}[X^2]\neq 0$.
