Estimating P(X $\ge$ k) with Chebyshev's inequality I have managed to derive non-rigorously that 

P(|X - E[X]| $\ge$ a) $\le$ $\frac{E[X - E[X]|^2}{a^2}$. for a random variable X.  

Now let X be a random variable with Poisson distribution, with mean and variance equal to $\lambda$. 
I am supposed to use this inequality to place an estimate on P(X $\ge$ k) where k $\ge$ 0. 

Initially I tried to do this by noting that P(X $\ge$ k) = 1 - P(X &lt; k) = 1 - P(0 $\le$ X &lt; k) and then writing this as an inequality, but this hasn't led anywhere. 

Any help would be greatly appreciated.
Chebyshev's inequality will be useful if $k \gg \lambda = E(X)$. In this case, you can write
$$
P(x \ge k) = P(X -\lambda \ge k-\lambda)\le \frac{E(X - \lambda)^2}{(k-\lambda^2)} = \frac{\lambda}{(k-\lambda)^2}, 
$$
which gives something nontrivial if $k \gg \lambda.$ If $k $ is much smaller than $\lambda$ then its not possible to get an upper bound which is useful. For example if $k=0$ then $P(X \ge 0) =1$. It would be helpful to know what kind of estimate you want exactly to give a more precise answer.
You want a more general inequality: $P(|X|\geq a)\leq E[|X|]/a$ or $P(|X|^k\geq a^k)\leq E[|X|^k]/a^k$. Chebyshev's inequality is a consequence of this.