Prove $P(X\geq1)\geq\frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$, assuming that $X$ has only natural values. I have to prove:
$$P(X\geq1)\geq\frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$$

Assuming that $X$ have only values from $\mathbb{N}$.

So, if I am correct we can write $\mathbb{E}X^2=\sum x_i^2p_i$ and $(\mathbb{E}X)^2=(\sum x_ip_i)^2$. If $X$ have only natural values, then $P(X\geq1)=1$

I think we should use Cauchy-Schwarz inequality here. So we start
$$
\frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}=\frac{(\sum x_ip_i)^2}{\sum x_i^2p_i}\leq\frac{(\sum x_i^2)(\sum p_i^2)}{\sum x_i^2p_i}
$$
And I do not know how to take it from here.
Solution

Since $Var(X) = \mathbb{E}X^2 - (\mathbb{E}X)^2 \geq 0$ 

We can write $\mathbb{E}X^2 \geq (\mathbb{E}X)^2$

Dividing both sides by $\mathbb{E}X^2$ gives:
$1 \geq \frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$

So $$P(X\geq1)\geq\frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$$
Solution

Since $Var(X) = \mathbb{E}X^2 - (\mathbb{E}X)^2 \geq 0$ 

We can write $\mathbb{E}X^2 \geq (\mathbb{E}X)^2$

Dividing both sides by $\mathbb{E}X^2$ gives:
$1 \geq \frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$

So $$P(X\geq1)\geq\frac{(\mathbb{E}X)^2}{\mathbb{E}X^2}$$