I am stuck in proving a property of a positive definite matrix.
Let $A$ be a positive definite matrix. Then: $$|A_{ij}| \le \sqrt{A_{ii}A_{jj}} \le \frac{A_{ii}+A_{jj}}{2}.$$
My work:
As for every non zero x, $x^TAx \gt 0$, consider $x^T = \begin{bmatrix}0& 0 &0 &\cdots& 0 &1& 0&\cdots&0 &-1& 0&\cdots&0 \end{bmatrix}$ with $1$ at position $i$ and $-1$ at position $j$. $$x^TAx = A_{ii}+A_{jj}-2*A_{ij} \gt 0.$$
Need help in completing the proof.
Just to expand Will Jagy's comment.
Consider $x^T= \begin{bmatrix} 0&\cdots &0& u&0&\cdots &0&v&0&\cdots&0\end{bmatrix}$, with $x_i^T = u$ and $x_j^T = v$. Following your thoughts, we can easily derive that: $$x^T A x = A_{ii}u^2 + 2A_{ij}uv + A_{jj}v^2 >0, \quad \forall u,v \in \mathbb R:u^2+v^2 \neq 0.$$
Now, let's fix a $v\neq 0$ and consider the function: $$f(u) = A_{ii}\cdot u^2 +2A_{ij}v\cdot u +A_{jj}v^2.$$
In order the function to have the same sign with the sign of $A_{ii}(> 0)$ for each $u\in \mathbb R$, the discriminant must be strictly negative, i.e. $$(2A_{ij}v)^2 - 4A_{ii}A_{jj}v^2 <0.$$ I guess you can take it from here.
The second inequality can be proven using the AM-GM inequality.
Even simpler, we want to prove $$\sqrt{A_{ii}A_{jj}} \le \frac{A_{ii} + A_{jj}}{2}\iff A_{ii} - 2\sqrt{A_{ii}A_{jj}} +A_{jj} \ge 0 \iff \left(\sqrt{A_{ii}} - \sqrt{A_{jj}}\right)^2\ge 0.$$ The last inequality apparently holds.
Another approach:
According to this Wikipedia article (property 3),
every principal submatrix of a positive definite matrix is positive definite.
Thus, if we consider the $2\times 2$ principal submatrix $A' = \begin{bmatrix} A_{ii} &A_{ij} \\ A_{ji} & A_{jj}\end{bmatrix}$ with $i \lt j$, we have that $$\det A'=A_{ii}A_{jj} - A_{ij}^2 \gt 0\implies A_{ij}^2 \lt A_{ii}A_{jj}\implies \lvert A_{ij}\rvert \lt \sqrt{A_{ii}A_{jj}}.$$