For some context about the inequality, it is in the proof of the Zygmund Paley inequality, I've posted another question on the same proof
I have a problem about the proof that asks to show that :
$$P(X \geq a ) \leq \frac{E[X^2]}{a^2+ E[X^2]}$$
assuming $E[X]=0$.
Which is okay so far, as the proofs I've read use the argument,
Then it says, let's assume now $E[X] \geq 0 $ only by applying the last inequality to another well chosen variable, conclude that:
$$P(X \geq E[X] + a ) \leq \frac{VarX}{a^2+ VarX}$$
(which by my trials I found it has to be by substituting the variable X by $X-E[X]$ )
$$P(X - E[X] \geq a ) \leq \frac{E[(X-E[X])^2]}{a^2+ E[(X-E[X])^2]}$$
which gives
$$P(X \geq E[X] + a ) \leq \frac{VarX}{a^2+ VarX}$$
My question here is more about the logic of this use in a proof, is it legitimate to generalize a result that was calculated for a variable whose expectation is null, but then generalized for a variable whose expectation again is not null,
by substituting $X$ with $E[X]+X$ , but if it is the case then we have to suppose the expectation of this new variable $E[X]+X$ to be null to be able to apply the result calculated for a variable whose expectation is null, that is $E[ E[X]+X ] =0$ , in which cases we get again $2E[X]=0$ so we have to suppose $E[X]=0$ and the generalization doesn't make sense for $E[X] \geq 0 $ only?
EDIT: I've made a mistake, the result is applied indeed to $E[X]-X$ and not $E[X]+X$, in this case should it make sense for $E[X] \geq 0 $ because we always have $E[ E[X]-X ] = 0 $?
Let it be that $\mathbb EX\in[0,\infty)$ and define $Y:=X-\mathbb EX$.
Then evidently $\mathbb EY=0$ so that we can conclude that:$$P(Y \geq a ) \leq \frac{\mathbb E[Y^2]}{a^2+ \mathbb E[Y^2]}$$
Since $Y=X-\mathbb EX$ this can also be expressed as:$$P(X-\mathbb EX \geq a ) \leq \frac{\mathbb E[(X-\mathbb EX)^2]}{a^2+ \mathbb E[(X-\mathbb EX)^2]}$$ or as:$$P(X\geq\mathbb EX + a ) \leq \frac{\mathsf{Var}X}{a^2+ \mathsf{Var}X}$$