Lower bound for probability of being above a certain percentage of the expectation

Question

Let $X$ be a non-negative random variable and $\lambda \in [0,1]$, can we show that $(1-\lambda)^2$ is smaller or equal to $P(X > \lambda E(X))$, where P denotes the probability distribution of $X$ and $E(X)$ its expectation?

Answer 1

No, take an arbitrary $n>0$ and a random variable with $P(X=n)=n^{-1}$, $P(X=0)=1-n^{-1}$, $P(X=a)=0$ for $a\notin \{0,n\}$. Then for any $\lambda>0$ the probability you're looking at is $n^{-1}$, which can be arbitrarily small.