How to prove the following inequality about $Var(x)$? Let,
$$ X = \frac{1}{n} \sum_{i=1}^n (h\left(s_i,p_i\right) - \mathbb{E}[h])z_i $$ where $(s_i,p_i)$ are input vectors and $z_i \leq B$. I am trying to proof
$Var(X) \leq |h\left(s_i,p_i\right)| \cdot |z_i| $ , Given that $E(X) = 0$.
My attempt
$$\begin{align}\mathsf{Var}(X) &= \mathsf E(~(X-\mathsf E(X))^2~) \\ &= \mathsf E(X^2-2X\mathsf E(X)+\mathsf E(X)^2) \\ &= \mathsf E(X^2)-2\mathsf E(X)\mathsf E(X)+\mathsf E(X)^2 \\ &= \mathsf E(X^2)-\mathsf E(X)^2\\ &= \mathsf E(X^2) \end{align}$$
Next how to prove my required inequality? Can one help me regarding this?

Edit.
I was trying to think in this way
Var(X) = deviation from the mean(X). So it is evident that it will be less than the highest possible value of that product of h and z. And E[h] is needless to say 0
Now can I write about that inequality directly?
 A: \begin{align}
\text{Var}(X) &= \text{E}[X^2] - (\text{E}[X])^2 \\
&= \text{E}[X^2] \\
&= \text{E} \left[\left(\frac{1}{n} \sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h]) z_i \right)^2\right] \\
&= \text{E} \left[\frac{1}{n^2} \left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h]) z_i \right)^2\right] \\
&\le \frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} |h(s_i,p_i) - \text{E}[h]| z_i \right)^2\right] \\
&\le \frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] \\
&= \frac{1}{n^2} A^2 B^2
\end{align}
Where we have used the Cauchy-Schwarz inequality in the fourth and fifth lines, and denoted $A^2 = \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right]$ and $B^2 = \text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right]$.
The Cauchy-Schwarz inequality states that for any two real-valued sequences of numbers ${a_i}$ and ${b_i}$, the following inequality holds:
\begin{equation}
\left(\sum_{i=1}^{n} a_i b_i \right)^2 \le \left(\sum_{i=1}^{n} a_i^2\right) \left(\sum_{i=1}^{n} b_i^2\right)
\end{equation}
In the fourth line of the proof, we use the Cauchy-Schwarz inequality by taking ${a_i} = |h(s_i,p_i) - \text{E}[h]|$ and ${b_i} = z_i$:
\begin{align}
\text{E} \left[\left(\sum_{i=1}^{n} |h(s_i,p_i) - \text{E}[h]| z_i \right)^2\right] &\le \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] \\
\end{align}
This inequality bounds the expected value of the square of the sum in the fourth line from above by the product of the expected values of the squares of the individual terms.
\begin{align}
\frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] &\le \frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] N \
&\le \frac{1}{n^2} M^2 N \
&= \frac{M^2 N}{n^2}
\end{align}
The first inequality follows from the fact that $\text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] \le N$ (since each $z_i^2$ is bounded above by $N$), and the second inequality follows from the fact that $\text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \le M^2$ (since each $|h(s_i,p_i)|$ is bounded above by $M$).
