Ratio of expectations for integer-valued random variable For a nonnegative integer-valued random variable $Y$ with positive expectation, show that $$\dfrac{E[Y]^2}{E[Y^2]}\leq\Pr[Y\neq 0].$$
I suppose that the probability that $Y=i$ is $x_i$, for $i=0,1,2,\ldots$. Then the inequality becomes
$$\dfrac{(x_1+2x_2+3x_3+\cdots)^2}{x_1+4x_2+9x_3+\cdots}\leq 1-x_1-x_2-\cdots$$ How can this be shown?
 A: So, following my earlier comment and your work so far, we have that our hypothesis is
$$\frac{(x_1 + 2x_2 + 3x_3 + \cdots)^2}{x_1 + 4x_2 + 9x_3 + \cdots}\leq x_1 + x_2 + x_3 + \cdots$$
multiplying both sides by the denominator (it is non-negative so the sign doesn't flip)
$$(x_1 + 2x_2 + 3x_3 + \cdots)^2 \leq (x_1 + x_2 + x_3 + \cdots)(x_1 + 4x_2 + 9x_3 + \cdots)$$
Looking at the $x_i x_j$ term in particular if $i\neq j$ you have on the left, $2ij x_i x_j$ and on the right $(i^2 + j^2) x_i x_j$  or if $i=j$ you have $i^2 x_i^2$ on the left and $i^2 x_i^2$ on the right which is always equal in the second case.
Looking at the first case more carefully, we have $2ij$ compared to $i^2 + j^2$.  Subtracting the left side, we have 0 compared to $i^2 - 2ij + j^2$ which is $(i-j)^2$ which is a real number squared and is always greater than or equal to zero.
Hence, since for every $i,j$ you have that the coefficient on the left is less than or equal to the coefficient on the right, the left side must be less than or equal to the right side, proving the statement.
A: If $E[Y] \gt 0$ then $\Pr(Y \ne 0) \gt 0$ and $E[Y] = E[Y\mid Y \not = 0]  \times \Pr(Y \not = 0) \gt 0$, 
so $E[Y]^2 = E[Y\mid Y \not = 0]^2  \times \Pr(Y \not = 0)^2 \gt 0$ and finite; 
and similarly $E[Y^2]  = E[Y^2\mid Y \not = 0]  \times \Pr(Y \not = 0) \gt 0$ possibly infinite.  
$Var(Y\mid Y \not = 0) = E[Y^2\mid Y \not = 0] - E[Y\mid Y \not = 0]^2$ is non-negative so $\dfrac{E[Y\mid Y \not = 0]^2}{E[Y^2\mid Y \not = 0]} \le 1$, 
so $\dfrac{E[Y]^2}{E[Y^2 ]} = \dfrac{E[Y\mid Y \not = 0]^2  \times \Pr(Y \not = 0)^2}{E[Y^2\mid Y \not = 0]  \times \Pr(Y \not = 0)} \le \Pr(Y \not = 0)$. 
A: By a comment of @HoCheol Shin, the previous answer had a mistake.

Corrected version:
By Titu's Lemma (a special case of the Cauchy-Schwarz inequality), we have that $$\frac{E[Y]^2}{E[Y^2]}=\frac{\left(\sum_{i=1}^\infty iP(Y=i)\right)^2}{\sum_{i=1}^\infty i^2P(Y=i)}\le \sum_{i=1}^\infty\frac{i^2P(Y=i)^2}{i^2P(Y=i)}=\sum_{i=1}^\infty P(Y=i)=P(Y\neq0)$$
