left-hand-tailed chebyshev (Cantelli's) inequality

#1
I'd like to use the one-tailed chebyshev inequality to help me describe some weather data. The proper inequality as I understand it is:
P{X-E(X)>=k(sigma)}<=1/(1+k^2)
Which describes the right-hand tail of the distribution (probability that values are above a bound).
Can someone help me prove to myself that a similar (or the same) equality exists for the left-hand tail (probability that values are below a lower bound)?
P{E(X)-X>=k(sigma)}<=???

Thanks-
Kevin

edit ... well, this is not homework. Still, any help?
 
#3
I should have mentioned that the definition is only for k>0.

What you're saying would be true if we were talking about |X-E(X)|.

Thanks for thinking through it with me.