Random variable $X\geq 0$ and its variance exists. How to prove


$$\mathbb{P}(X\geq(1-t)\mathbb{E}(X))\geq \frac{t^2\mathbb{E}(X)^2}{\mathbb{E}(X^2)}\enspace\text{for}\enspace t\in(0,1]$$


$$\mathbb{E}(\exp(-\lambda(X-\mathbb{E}(X))))\geq\exp(\lambda^2\mathbb{E}(X^2)/2)\enspace\text{for}\enspace \lambda\geq0$$


$$\mathbb{P}(X\leq(1-t)\mathbb{E}(X))\leq \exp(-\frac{t^2\mathbb{E}(X)^2}{2\mathbb{E}(X^2)})\enspace\text{for}\enspace t\in(0,1]$$


Thanks!




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Update: Thanks to [Alapan Das](https://mathoverflow.net/users/156029/alapan-das) and [Iosif Pinelis](https://mathoverflow.net/users/36721/iosif-pinelis) I found that the first inequality is Paley-Zygumund Inequality. It can be transformed to the shape similar to the third equation i.e.
$$\mathbb{P}(X\leq(1-t)\mathbb{E}(X))\leq 1-\frac{t^2\mathbb{E}(X)^2}{\mathbb{E}(X^2)}\enspace\text{for}\enspace t\in(0,1]$$
but the third one is a tighter bound. I wonder how to improve it.



Prove inequality on expectation
As Alapan Das noted, the first inequality for $t\in[0,1]$ is the [Paley–Zygmund inequality](https://en.m.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality). For $t<0$ or $t>1$, the first inequality is false if e.g. $X=1$.


The second inequality is false for all real $\lambda\ne0$ if e.g. $X=1$.


The third inequality is false for all real $t>0$ if e.g. $X=1$. It is also false for all real $t\le-2$ if e.g. $P(X=1)=\frac1{1-t}=1-P(X=0)$.




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The OP has changed the 3rd inequality from


$$P(X\ge(1-t)EX)\ge\exp-\frac{t^2\,(EX)^2}{2EX^2}$$
for an unspecified $t$ to
$$p\_3:=P(X\le(1-t)EX)
\le\exp-\frac{t^2\,(EX)^2}{2EX^2}\tag{3a}\label{3a}$$
for $t\in(0,1]$,
at the same time asserting that the upper bound in \eqref{3a} on $p\_3$ is "tighter" than the Paley–Zygmund bound
$$1-\frac{t^2\,(EX)^2}{EX^2}$$
on $p\_3$. In fact, it is vice versa:
$$1-\frac{t^2\,(EX)^2}{EX^2}<\exp-\frac{t^2\,(EX)^2}{EX^2}
<\exp-\frac{t^2\,(EX)^2}{2EX^2}$$
(if $EX^2\ne0$).


