Is there one-tailed version of Vysochanskiï–Petunin inequality, like Chebyshev? The Vysochanskiï–Petunin inequality gives a tighter bound than Chebyshev for unimodal distributions . I'm just wondering if there is a one tailed version of it, like that of Chebyshev inequality? Please help.
 A: We are talking about continuous random variables with unimodal densities. The Wikipedia article you link to says 

for any $ \lambda \gt \sqrt{8/3}=1.63299\ldots$, you have
  $\Pr(\left|X-\mu\right|\geq \lambda\sigma)\leq\dfrac{4}{9\lambda^2}.$

This, which I wrote many years ago and have not checked recently, says (trying to translate to a similar notation)

Unimodal two-tailed case:
  
  
*
  
*If $\lambda \ge  B  $, then $\Pr(|X-\mu|\ge \lambda\sigma) \le \dfrac{4 }{\; 9 \lambda^2}$
  
*If $\lambda \le  B  $, then $\Pr(|X-\mu|\ge \lambda\sigma) \le 1-\left(\dfrac{4 \lambda^2}{3(1+\lambda^2)}\right)^2$
where B is the largest root of $7x^6-14x^4-x^2+4=0$, about $1.38539\ldots...$

which seems similar but slightly stronger.
It also says 

Unimodal one-tailed case:
  
  
*
  
*$\Pr(X-\mu \ge \lambda\sigma) \le \max \left\{ \dfrac{4}{9(1+\lambda^2)}, \dfrac{3-\lambda^2}{3(1+\lambda^2)}  \right\}$

so taking the first term if $\lambda \ge  \sqrt{\frac{5}{3}}$ and the second if $0 \le \lambda \le  \sqrt{\frac{5}{3}}$
