$P(\mathbf{X} \geq t) \leq P((\mathbf{X} + a)^2 \geq (t+a)^2))$ In general, is it true that
$$ P(\mathbf{X} \geq t) \leq P((\mathbf{X} + a)^2 \geq (t+a)^2))$$ for non-negative $a$.
What about the general case of replacing square with a function $\Phi(\mathbf{X})$, where $\Phi$ is a non-decreasing non-negative function. 
How can I prove this ?
 A: In general, inequalities of probabilities involve a relationship between the events involved. Thus, if $A,B$ are "events" (i.e., measurable sets with respect to the underlying probability measure) such that $A\subset B$, then $P(A)\leq P(B)$ and the reason is self-evident: larger sets have larger probability. Now, if you think about $X\geq t$ as defining an "event" in your probability space, and $\Phi$ is a non-decreasing positive function, then what you have to check in order to prove your inequality is that the event $X\geq t$ is contained in the event $\Phi(X)\geq \Phi(t)$. That is, whenever $X\geq t$, then also $\Phi(X)\geq \Phi(t)$. Since this holds for every non decreasing function, you get your proof.
A: No, it is not true. Take $X=0$, $t=-1$, $a=1$. Then the left hand side is $1$ and the right hand side is $0$.
In general, we do have $P(X\ge t)=P(X+a\ge t+a)$ (for any $a$, not just non-negative) but we do not have $P(X\ge t)\le P(X^2\ge t^2)$. This will hold if $X$ is non-negative, however.
