We have mapping $ \left \langle x,x \right \rangle = x^{T}Ax $. Prove that $ \left \langle x,x \right \rangle > 0 $ iff... Let's suppose we have the mapping $ \left \langle x,y \right \rangle = x^{T}Ay $ ,
A is the symmetric matrix  $\begin{pmatrix}
a & b\\ 
b & c
\end{pmatrix} $, $a,c >0$
And I need to proof that $ \left \langle x,y \right \rangle$ is a scalar product iff $ac-b ^{2} > 0$.
I've already proved that this mapping is linear and $\left \langle x,y \right \rangle = \left \langle y,x \right \rangle $ for every $x$ and $y$.
Now I want to show that $ \left \langle x,x \right \rangle > 0 $ iff  $ac-b ^{2} > 0$,
   for $ x\neq 0 $ .
Let's suppose that $x=\binom{x_1}{x_2}$
So I came to this 
$$x_{1}^{2}a + 2x_{1}x_{2}b + x_{2}^{2}c >  0 $$
$$x_{1}^{2}a + b^{2}\left \langle x,x \right \rangle  + x_{2}^{2}c >  0 .$$
Now I see that $x_{1}^{2}+x_{2}^{2}= \left \| x \right \|^{2}$ but I stack here, because I don't know how to use this fact.

As mentioned in the comments and the other answer, this problem can easily be answered using eigenvalues of $A$.
However, if you don't know what eigenvalues are, you can simply multiply the inequality $x_{1}^{2}a + 2x_{1}x_{2}b + x_{2}^{2}c >  0$ by $a>0$, which gives you
$$x_{1}^{2}a^2 + 2x_{1}x_{2}ab + x_{2}^{2}ac >  0.$$
Now rewrite the right side as
$$x_{1}^{2}a^2 + 2x_{1}x_{2}ab +x_2^2b^2+ x_{2}^{2}(ac-b^2)=(ax_1+bx_2)^2+(ac-b^2)x_2^2.$$
Now if $ac-b^2\leq 0$, this can be nonpositive for $(x_1,x_2)=(b,-a)\neq (0,0)$. On the other hand, if $ac-b^2>0$, then it is nonnegative, and can only be $0$ when $x_2=x_1=0$.
