# Using Probabilistic argument to show an inequality

Question:-Use Chebychev's inequality to show that for any $k>1$, $e^{k+1}\ge k^2$

Chebychev's inequality states that for any random variable $X$ with finite mean, and for $k>0$ $$P(|X-\mu|\le k\sigma)\ge 1-\frac {1}{k^2}$$ Or $$P(|X-\mu|\ge k\sigma)\le \frac {1}{k^2}$$ A general for of the inequality is given in the book of Casella-Berger, $$P(h(X)\ge\epsilon)\le\frac{Eh(X)}{\epsilon}$$ I tried to do it using the argument that for $k>1$ we have to show $P(k^2\le e^{k+1})\ge$ a no. quite close to $1$.

However,I am getting problem in thinking a random variable which will serve the work. I can't even consider the function as $h(k)$ because both the function on either side of the inequality contains $k$.

Hoping to get a hint by my fellow mates.

Set $X\sim \exp(1)$. Then you have for $k>1$ \begin{align} e^{-(k+1)}=P(X>1+k) = P(|X-1|>k)\leq \frac{1}{k^2} \end{align} And now it is easy to finish.

• Thank you very much Shashi. – user440191 Feb 1 '18 at 1:53