Q:

Let $a,b,c$ are positive real numbers such that $b^2+c^2<a<1.$

I came across the following question that says:  

Let $a,b,c$ are positive real numbers such that $b^2+c^2<a<1.$ Consider the $3 \times 3$ matrix $A=\begin{pmatrix}
1 &b  &c \\ 
b &a  &0 \\ 
c &0  &1 
\end{pmatrix}$.  Now I have to show that all the eigenvalues of $A$ are positive real numbers.

My Attempt: I have taken the column matrix $z=\begin{pmatrix}
x &y  &z 
\end{pmatrix}^T$. Now,$z^TAz=x^2+ay^2+z^2+2bxy+2czx.$ I have to show that $z^TAz>0.$ The information I have on my hand is: $a,b,c >0,b^2+c^2<a<1$ and $A$ is symmetric . Now how can I make a conclusion from these that  $z^TAz>0 ?$
Can someone point me in the right direction? Thanks in advance for your time.  
EDIT: As @DonAntonio commented that $A$ is positive definite symmetric matrix and so all the eigenvalues of $A$ are positive real numbers. But for this I have to show that $z^TAz>0 .$ How Can I do that? Do I have to find out the eigenvalues of $A$ independently and then make  a conclusion from it? I am still confused.

A:

For symmetry: 
$$A=\begin{pmatrix}
1 &b  &c \\ 
b &a  &0 \\ 
c &0  &1 
\end{pmatrix}$$ $\Longrightarrow $ 
$$
A^{T} = \begin{pmatrix}
1 &b  &c \\ 
b &a  &0 \\ 
c &0  &1 
\end{pmatrix} = A  \; \; \; \checkmark 
$$
For positive definiteness notice that $A$'s leading principal minors are: 
$$
1>0 \; \; \; \checkmark \\
a-b^2> c^2 > 0 \; \; \; \checkmark  \\
a + -b\cdot(b) + c\cdot(-ac) = a-b^2 - ac^2> 0 \; \; \; \checkmark 
$$
Note that a symmetric matrix is just a special case of a hermitian matrix.
A hermitian matrix is positive definite if its leading principal minors are all positive. Indeed we observe this, so that the matrix $A$ is positive definite symmetric. Thus the matrix $A$ has all real positive eigenvalues.  
Definition
The $k^{\text{th}}$ leading principal minor of a matrix $M$ is the determinant of its upper left $k$ by $k$ sub-matrix.

