Chebyshev's inequality states that
for t>0 
A onetailed version of Chebyshev's
inequality is that for t>0 
or go to a ModeMean inequality (and a clearer pdf) or ModeMedianMean relationships or some Statistics Jokes 
<= becomes = when distribution X is (for
given k>=1 and any m and any s>0)
P[X=mk.s] = 1/(2.k^{2}), P[X=m+k.s] = 1/(2.k^{2}),
and P[X=m] = 11/k^{2} .
Note E(X)=m and Var (X)=s^{2}, sd(X)=s, so for this
X P[XE(X)>=k.sd(X)] = 1/k^{2} .
The equality will in general only be achieved for a symmetric
threevalued distribution. If the probabilities are p, 12p and p
then equality is achieved when k=(2p)^{1/2}. A symmetric
twovalued distribution is a special case with k=1.
A chart showing this distribution for k=2 is below.
Note also that for 1>k>0, 1/k^{2}>1 which is not
a probability, so achieving equality would be impossible.
<= becomes = when distribution X is (for
given k>0 and any m and any s>0)
P[X=m+s.k] = 1/(1+k^{2}), P[X=ms/k] = k^{2}/(1+k^{2}) .
Note E(X)=m and Var (X)=s^{2}, sd(X)=s, so for this
X P[XE(X)>=k.sd(X)] = 1/(1+k^{2}) .
The equality will in general only be achieved for a twovalued
distribution. If the probability of the higher value is p (and
the lower 1p) then equality is achieved when k=(1/p1)^{1/2}
A chart showing this distribution for k=2 is below.
Note also that for all k>0, 1>1/(1+k^{2})>0 so,
unlike the twotailed version, this is always meaningful.
Markov's inquality states that for Y nonnegative and s>0
P[Y>=s] <= E(Y)/s .
If Y=(XE(X))^{2} and s=t^{2} with t>0 then
P[(XE(X))^{2}>=t^{2}] <= E(XE(X))^{2}/t^{2}
so P[XE(X)>=t] <= Var(X)/t^{2}
.
If t^{2}=k^{2}.Var(X) for k>0 then k.sd(X)=t
and P[XE(X)>=k.sd(X)] <= 1/k^{2} .
If P[XE(X)>=t] = 0 for given t>0 then clearly
true,
otherwise let a = P[XE(X)>=t], noting 0<a<1,
and b = E(XE(X)XE(X)>=t), noting t<=b,
and c = E((XE(X))^{2}XE(X)>=t), noting t^{2}<=c<Var(X)/a
Consider T=t if XE(X)>=t and T=0 if XE(X)<t and
consider S=0 if XE(X)>=t and S=(XE(X)).t/b if XE(X)<t :
then E(T) = a.t and E(S) = a.b.t/b = a.t
and E(SXE(X)<t) = a.t/(1a)
and
E(T^{2}) = a.t^{2} <= a.c
and E(S^{2}) = (Var(X)a.c).t^{2}/b^{2} <= Var(X)a.c ,
so Var(X) >= E(T^{2})+E(S^{2}) ;
but E(S^{2}) = (1a).E(S^{2}XE(X)<t) >= (1a).(E(SXE(X)<t))^{2}=a^{2}.t^{2}/(1a)
so Var(X ) >= a.t^{2}+a^{2}.t^{2}/(1a) = a.t^{2}/(1a) ,
so a <= 1/(1+t^{2}/Var(X))
so P[XE(X)>=t] <= 1/(1+t^{2}/Var(X))
i.e. P[XE(X)>=t] <= Var(X)/(Var(X)+t^{2}) .
If t^{2}=k^{2}.Var(X) for k>0 then k.sd(X)=t
and P[XE(X)>=k.sd(X)] <= 1/(1+k^{2}) .
The proof above was mine. A.G.McDowell has two at his page on Chebyshev's Inequalities: one of his own and one from 'Probability and Random Processes', by Grimmett and Stirzaker, published by Oxford Science Publications, ISBN 0 19 853665 8. The latter has the advantage of being short and (slightly adapted) is something like this:
With t>0, for any c>=0 P[XE(X)>=t] = P[XE(X)+c>=t+c]
<= E((XE(X)+c)^{2}/(t+c)^{2}) = (Var(X)+c^{2})/(t+c)^{2}.
But treating (Var(X)+c^{2})/(t+c)^{2} as a
function of c, the minimum occurs at c = Var(X)/t,
so P[XE(X)>=t] <= (Var(X)+Var(X)^{2}/t^{2})/(t+Var(x)/t)^{2}
= Var(X)/(t^{2} + Var(x)).
In 2001 I produced yet another proof.
Pafnuty Lvovich Chebyshev was a notable Russian mathematician,
who was born on 16 May 1821 and died on 8 December 1894. He wrote
on number theory, analysis, probability, mechanics and maps.
Being Russian, his name was written in the Cyrillic alphabet,
causing problems of translitteration. Stijn van
Dongen has compiled a list of 33 western European verions of
his surname, of which Chebyshev seems to be the most popular in
English, Tchebycheff in French, Cebysev in Spanish, and
Tchebyscheff in German. Despited this variety, the final vowel
always seems to be transcribed as "e" in that list when
it is in fact pronounced "yo" in Russian. Even his
given name and his patronimic have varying versions in roman
script.
The two inequalities are very similar in form. However, for small k, they produce very different results. Indeed the onetailed version produces meaningful results for 0<k<1, while Chebyshev's inequality less helpfully limits the probability to being less than or equal to a number greater than 1. For k=1, the onetailed version provides the result that the median of a distribution is within one standard deviation of the mean. For large k, as shown in the table and chart below, the mamimum probability values given by the two inequalities are very similar. For large k, they are very different to the results for the Normal distribution.
P(XE(X)>=k.sd(X)) or P(XE(X)>=k.sd(X)) 

k  Chebyshev's inequality 
One tailed version 
Normal twotailed 
Normal onetailed 

0.1  (10000%)  99%  92.0%  46.0%  
0.5  (400%)  80%  61.7%  30.9%  
0.9  (123%)  55%  36.8%  18.4%  
1  100%  50%  31.7%  15.9%  
1.64  37.0%  27.0%  10%  5%  
1.96  26.0%  20.7%  5%  2.5%  
2  25%  20%  4.55%  2.28%  
3  11.11%  10%  0.27%  0.13%  
4  6.25%  5.88%  0.006%  0.003%  
4.36  5.26%  5%  0.0013%  0.0007%  
4.47  5%  4.76%  0.0008%  0.0004%  
9.95  1.01%  1%  close to 0  close to 0  
10  1%  0.99%  close to 0  close to 0 
The similarity of the probability values for high k between Chebyshev's inequality and the onetailed version is slightly misleading, as can be seen when the distribributions which turn the inequalities into equalities are considered, with examples in the chart below. The equality distribution for Chebychev's inequality is symmetric, while the equality distribution for the onetailed version is not. For large k, the highest element of the former has only just over half the probability of the highest element of the latter.
Commenting in sci.stat.math, Herman Rubin, Department of Statisitics, Purdue University, said:
"This goes back more than a century. I do not know the history.
"This is the simplest of a class of inequalities,
which I believe are due to Markov, a student of Chebyshev. A good
book on the moment problem should have a reference.
If 1=m_0, m_1, ..., m_{2n} are the first 2n moments of a
probability distribution,
let f be the unique (apart from a constant of proportionality)
polynomial of degree at most n+1 with x as a root, and orthogonal
to all polynomials of degree less than n.
If x is not a root of the orthogonal polynomial of degree n,
there are n+1 roots, and the unique distribution concentrated at
these roots fitting the given moments maximizes P(X <= x) and
minimizes P(X < x).
If x is a root of the orthogonal polynomial, the 2nth moment may
not be matched at the n roots.
"In the case n=1, If x=E(X), the extreme distribution
is concentrated at E(X).
Otherwise, if x = E(X)+t, the distribution is concentrated at x
and E(X)Var(X)/t."
Copyright June 1999 Henry Bottomley. All rights reserved.
If X is a continuous random variable with a unimodal probability density function (pdf), we may be able to tighten Chebyshev's inequality, though only by adding some complexity.
If the unimodal probability density function is
also symmetric, then result
P[XE(X)>=t] <= 4.Var(X)/(9.t^{2})
or P[XE(X)>=k.sd(X)] <= 4/(9.k^{2})
applies for all t,k>0
This is Gauss's inquality of 1821 when the mode and mean are
identical,
and is four ninths of the bound of Chebshev's inequality.
When 0<t^{2}<=4.Var(X)/3
or 0<k<=2/sqrt(3) about 1.15470...
I think this can be improved to
P[XE(X)>=t] <= 1t/sqrt(3.Var(X)) or P[XE(X)>=k.sd(X)] <= 1k/sqrt(3)
It is obvious that for a symmetric pdf, the onesided bounds
are half these.
If the unimodal pdf need not be symmetric then I
suspect the inequality only applies for high t and k:
when t^{2}>=B^{2}.Var(X)
or k>=B
where B is the largest root of 7.x^{6}14.x^{4}x^{2}+4=0,
about 1.38539...
P[XE(X)>=t] <= 4.Var(X)/(9.t^{2})
or P[XE(X)>=k.sd(X)] <= 4/(9.k^{2})
as before,
but when 0<t^{2}<=B^{2}.Var(X)
or 0<k<=B and with B as before,
about 1.38539...
P[XE(X)>=t] <= 1(4.t^{2}/(3.(Var(X)+t^{2})))^{2}
or P[XE(X)>=k.sd(X)] <= 1(4.k^{2}/(3.(1+k^{2})))^{2}
(I produced a sketch
of why this is true in 2002)
The one sided version might be:
when t^{2}>=5.Var(X)/3
or k>=sqrt(5/3), about 1.29099...
P[XE(X)>=t] <= 4.Var(X)/(9.(Var(X)+t^{2}))
or P[XE(X)>=k.sd(X)] <= 4/(9.(1+k^{2}))
this time four ninths of the one sided version of Chebyshev's
inequality
and when 0<t^{2}<=5.Var(X)/3
or 0<k<=sqrt(5/3)
P[XE(X)>=t] <= 14.t^{2}/(3.(Var(X)+t^{2}))
or P[XE(X)>=k.sd(X)] <= 14.k^{2}/(3.(1+k^{2}))
(I have attempted a proof
of this in 2001)
In an email in October 2000, Michael Dummer commented:
"there is a simple unimodal distribution which meets the
tighter Unimodal Chebyshev bound: a rectangular or uniform
distribution with an impulse at the midpoint of the uniform range."
I responded:
"I think this applies to the symmetric unimodal case; for
the nonsymmetric unimodal two and onesided cases, the bound is
met for uniform distributions with an impulse away from the
midpoint." Perhaps I should have added the bounds for small t
and k are achieved for a uniform distribution
with an impulse at one end where the probability of the uniform
distribution is 4.k^{2}/(3.(1+k^{2})).
I have developed a page which takes the onetailed case a little further (especially the unimodal case) as well as some medianmeanmode inequalities.