<s>[INST] How can I derive the conditions of Positive semidefinite cone in $2\times2$ matrix. By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that
\begin{equation}
X=\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge0, \quad xz\ge y^2,\tag{1}
\end{equation}
where
$$
S_+^2 = \left\{X\in S^2 | X \succeq 0\right\}.
$$

I have failed to prove the $(1)$.

$$
\begin{align}
\forall \alpha,\beta,\quad 
\left[
\begin{array}{cc}
\alpha & \beta
\end{array}
\right]
\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]
\left[
\begin{array}{cc}
\alpha \\
\beta
\end{array}
\right] &\ge 0\\
\alpha^2x + 2\alpha\beta y + \beta^2z &\ge 0\\
\left(\alpha\sqrt{x} + \beta\sqrt{z}\right)^2 + 2\alpha\beta(y-\sqrt{xz}) &\ge 0
\\\therefore \alpha\beta(y-\sqrt{xz})\ge0\\
\text{if}\quad\alpha\beta\ge0,\quad \text{then} \quad y^2\ge xz\\
\text{if}\quad\alpha\beta\le0,\quad \text{then} \quad y^2\le xz
\end{align}
$$

How can I reach to 
$$
x\ge0,\quad z\ge0,\quad xz\ge y^2\qquad?
$$
 [/INST] This is essentially the implication from the Sylvester's criterion extended to the case of positive semi-definite matrices.
All of the leading principal minors must be nonnegative. So you immediately get $x\ge 0, x z\ge y^2, z\ge 0$.
Following your proof for $2\times2$ matrices, take consequently $\alpha = 0$ and $\beta = 0$ to get $x\ge 0, z\ge 0$. The last step is to take $\alpha = \beta = 1$
$$2y \ge -x-z\Rightarrow 4y^2 \ge (x+z)^2 \ge 4xz$$
</s>