$\def\Var{\mathop{\rm Var}} \def\E{{\mathrm E}\,} $
X is non-negative r.v.
Since, $\Var[X]$ is a sum of square so I knew that $$\Var[X] \geq 0$$
I have to prove that $\Var[X] \geq 0$ by using Cauchy-Schwarz Inequality.
I did prove this $$\E[XY]^2 \leq \mathrm \E[X^2] \E[Y^2]$$ by quadratic equation.
But, I still have no idea on proving this problem.
I did try to multiple $\frac{(1^2+1^2+…+1^2)}{n}$ on the $\Var[X] = \E[(X-\E[X])^2$
However, I realize that in the Cauchy-Schwarz, it also need $a_i = cb_i$ to make it true. So, I could not assume this.
I also tried to read on wikipedia about this inequality
but there is nothing about this step.
May anyone guide me anything to start with?
Best Answer
Its just that $Var(X)=E((X-E(X))^2)$ which are $(X-E(X))^2$non-negative random variables. Expectation of non-negative random variables is non-negative.