Suppose $\mathbb{E}(X)=0 , \text{Var}(X)=1$ and $|X|\leq M$ on $\Omega$.
Prove that $$P(|X|\geq t)\geq \frac{1-t^2}{M^2-t^2} \quad\forall\,0<t<1.$$
How I can prove , I attempted to use Markov's inequality but this does not hold.
Please help me to prove this, thanks in advance.
You need a slight strengthening of the $g(x)=x^2$ instance of the "Basic inequality" from Loève's Probability Theory $\S$9.3 A, but you can get it with a small modification of the proof there. Letting $f_X$ be the density of $X$, $$ 1=E(X^2) = \int_{|x|\geq t} x^2 f_X(x) dx + \int _{|x|<t} x^2f_X(x) dx $$ Because $x^2$ is increasing, the second integral is bounded above by $t^2 P(|X|<t) = t^2(1-P(|X|\geq t))$ and the first is at most $M^2 P(|X|\geq t)$. Therefore $$E(X^2) \leq (M^2-t^2)P(|X|\geq t) + t^2 $$ from which your inequality follows.