Showing that given matrix does not have negative eigenvalues without using the knowledge that it is positive definite.
The characteristic polynomial of your matrix is $$ p(x) = (a-x)(1-x)^2-c^2(a-x)-b^2(1-x). $$ Now, if $x < 0$, then \begin{align*} p(x) &> (a-x)(1-x)^2-c^2(1-x)-b^2(1-x) = (1-x)\cdot[(a-x)(1-x)-c^2-b^2]\\ &> (1-x)\cdot[x^2-(a+1)x] = x(1-x)(x-a-1) > 0. \end{align*} Therefore, $p$ cannot have zeros in $(-\infty,0)$. Also, $p(0) = \det A > 0$. Thus, the eigenvalues of $A$ are positive.
You may use Sylvester's Law of Inertia. Here is a matrix of determinant $1$
$$ R = \left( \begin{array}{ccc} 1& -b & \frac{-ac}{a-b^2} \\ 0&1 & \frac{bc}{a-b^2} \\ 0&0 &1 \\ \end{array} \right) $$ and a "congruence diagonalization" $R^T AR = D$ $$ \left( \begin{array}{ccc} 1&0 &0 \\ -b&1 &0 \\ \frac{-ac}{a-b^2}& \frac{bc}{a-b^2}&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1&b &c \\ b& a&0 \\ c& 0&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1& -b & \frac{-ac}{a-b^2} \\ 0&1 & \frac{bc}{a-b^2} \\ 0&0 &1 \\ \end{array} \right) = \left( \begin{array}{ccc} 1&0 &0 \\ 0&a-b^2 &0 \\ 0&0 & \frac{a-ac^2 - b^2}{a-b^2}\\ \end{array} \right) $$
I found $R$ using a fairly clean algorithm, see http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
Let's see, I have written a C++ program for this congruence diagonalization, but it accepts only matrices with integer entries (and square and symmetric). In my program, the matrix of determinant $1$ uses only rational entries. For your problem, with symbol entries, back to one step at a time. However, it is exactly the same algorithm. The bottom line is that you do not need to calculate the eigenvalues themselves, this process tells you how many are positive, how many are exactly zero, how many are negative.
parisize = 4000000, primelimit = 500000
? h = [ 1,b,c; b,a,0; c,0,1]
%1 =
[1 b c]
[b a 0]
[c 0 1]
? ht = mattranspose(h)
%2 =
[1 b c]
[b a 0]
[c 0 1]
? h-ht
%3 =
[0 0 0]
[0 0 0]
[0 0 0]
? r1 = [ 1,-b,-c; 0,1,0; 0,0,1]
%4 =
[1 -b -c]
[0 1 0]
[0 0 1]
? r1t = mattranspose(r1)
%5 =
[ 1 0 0]
[-b 1 0]
[-c 0 1]
? r1t * h * r1
%6 =
[1 0 0]
[0 -b^2 + a -c*b]
[0 -c*b -c^2 + 1]
? r2 = [ 1,0,0; 0,1,(b * c)/(a - b^2); 0,0,1]
%7 =
[1 0 0]
[0 1 c*b/(-b^2 + a)]
[0 0 1]
? r2t = mattranspose(r2)
%8 =
[1 0 0]
[0 1 0]
[0 c*b/(-b^2 + a) 1]
? r2t * r1t * h * r1 * r2
%9 =
[1 0 0]
[0 -b^2 + a 0]
[0 0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]
? r
%10 = r
? r = r1 * r2
%11 =
[1 -b -a*c/(-b^2 + a)]
[0 1 c*b/(-b^2 + a)]
[0 0 1]
? rt = mattranspose(r)
%12 =
[ 1 0 0]
[ -b 1 0]
[-a*c/(-b^2 + a) c*b/(-b^2 + a) 1]
? h
%13 =
[1 b c]
[b a 0]
[c 0 1]
? r
%14 =
[1 -b -a*c/(-b^2 + a)]
[0 1 c*b/(-b^2 + a)]
[0 0 1]
? rt * h * r
%15 =
[1 0 0]
[0 -b^2 + a 0]
[0 0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]
? d = rt * h * r
%16 =
[1 0 0]
[0 -b^2 + a 0]
[0 0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]
? q = d[3,3]
%17 = (-b^2 + (-a*c^2 + a))/(-b^2 + a)
? p
%18 = p
? p = q * (a - b^2)
%19 = -b^2 + (-a*c^2 + a)
?