Let $a>b>0$, $0<p<1$ and let $X$ have $P(X=a)=p$ and $P(X=-b)=1-p$. Apply Chebyshev inequality to $\varphi (x)=(x+b)^2$ and conclude that if $Y$ is any random variable with $E(Y)=E(X)$ and $Var(Y)=Var(X)$ then $P(Y\geq a)\leq p$ and equality holds if and only if $Y=X$.
The first part is easy. But I stuck in the "Equality part". I know that to show to r.v. equal a.s. all you need to show is to show the equality in M.G.F., but here we can not have the M.G.F. of $Y$.
Note: Chebyshev’s inequality :- Suppose $\varphi : \mathbb{R}\rightarrow \mathbb{R}$ has $\varphi \geq 0$ let $A\in \mathcal{B}$ (Borel set) and let $i_A = \inf \{ \varphi (y) :y\in A \}$.
$i_A P(X\in A) \leq E(\varphi (X), X\in A)\leq E\varphi (X)$
SOLUTION OF THE INEQUALITY:
Since $E(Y)=E(X)$ and $Var(Y)=Var(X)$ then $E(Y^2)=E(X^2)$
$i_A P(Y\geq a)\leq E(Y+b)^2=E(X+b)^2=p(a+b)^2$. Here $A=\{ Y\geq a \}$. Then $i_A=(a+b)^2$
So $P(Y\geq a)\leq p\space \space\space\space \blacksquare$
In the equality case, all the long chain of inequalities becomes a chain of equalities. In particular, $E((Y+b)^2)=E((Y+b)^21_A)=p(a+b)^2$. It follows that $E((Y+b)^2 1_{A^{c}})=0$ where $A^c$ denotes the complement of $A$. The variable $Z=(Y+b)^2 1_{A^{c}}$ is nonnegative with mean $0$, so it must be zero a.s. In other words, we have $(Y+b)^2=0$ (i.e. $Y=-b$) outside of $A$.
Similarly, we have $E((Y+b)^21_A)=E((a+b)^21_A)$ so $E((Y-a)(Y+a+2b)1_A)=0$. The variable $T=(Y-a)(Y+a+2b)1_A$ is nonnegative with mean $0$, so it must be zero a.s. In other words, we have $(Y-a)(Y+a+2b)=0$ (i.e. $Y=a$) on $A$. This finishes the proof.