$A\in\mathbb{R}^{n\times n}$ is symmetric and positive definite $A\in\mathbb{R}^{n\times n}$ is symetric and positive definite.
Show that $\forall i\neq j; i,j=1,2,...,n; $   $a^2_{ij}<a_{ii}a_{jj}$
I tried to do this by contradiction and show that one of the leading principal minors isn't positive but this didn't lead me anywhere and I'm not sure it is a good way.

Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb{R}^n$. 
Notice that $(e_i+\lambda e_j)^tA(e_i+\lambda e_j)=\lambda ^2a_{jj}+2\lambda a_{ij}+a_{ii}>0$, for every $\lambda\in\mathbb{R}$.
Therefore $(2a_{ij})^2-4a_{jj}a_{ii}<0$.
