The Chebyshev's inequality is $P(|X-E(X)|>\varepsilon)\leq \frac{\operatorname{Var(X)}}{\varepsilon^2}.$
I saw a proof which goes like this:
$ \begin{align} \operatorname{Var(X)}(X) &= E((X-E(X))^2) \\ &= \sum_{x\in S}(x-E(X))^2\cdot P(X=x) \\ &\geq \sum_{|x-E(X)|>\varepsilon}(x-E(X))^2\cdot P(X=x) \\ &> \sum_{|x-E(X)|>\varepsilon}\varepsilon^2\cdot P(X=x) \\ &= \varepsilon^2 P(|X-E(X)|>\varepsilon) \\ \end{align} $
from which the equation should follow by dividing by $\varepsilon^2$.
What I don't understand here is the 4th step:
$\sum_{|x-E(X)|>\varepsilon}(x-E(X))^2\cdot P(X=x) > \sum_{|x-E(X)|>\varepsilon}\varepsilon^2\cdot P(X=x) $
Doen't this imply $P(|X-E(X)|>\varepsilon)< \frac{\operatorname{Var(X)}}{\varepsilon^2}$ rather then $P(|X-E(X)|>\varepsilon)\leq \frac{\operatorname{Var(X)}}{\varepsilon^2}.$
Why is this correct?