need help with proving that a matrix has at least one positive eigenvalue I need to prove, that the matrix
$ \begin{pmatrix}
A &B \\ 
B & C
\end{pmatrix} $ has at least one positive eigenvalue, if known that $ A+4B+5C > 0 $.
I was told to show that $ \begin{pmatrix}
1 &2 \\ 
2 & 5
\end{pmatrix} $ is positive definite. But I don't know what to do with that hint, and what is the connection between the inequality and and the matrix.
Two things I know and think that might be usefull is that the sum of eigenvalues is the trace of the matrix, and their product is the determinent
any usefull hints/directions?
big thank you
 A: Here's a way to do it without the hint. 
The characteristic polynomial of the matrix is $$\lambda^2-(A+C)\lambda+AC-B^2\tag1$$ so the eigenvalues are $${A+C\pm\sqrt{(A-C)^2+4B^2}\over2}\tag2$$ First, note that what's under the square root sign is non-negative, so the eigenvalues are real (this could also have been deduced from the theorem that says that the eigenvalues of a symmetric matrix are real). The larger eigenvalue is the one with the plus sign, so we just have to prove that if $A+4B+5C\gt0$ then $$A+C+\sqrt{(A-C)^2+4B^2}\gt0\tag3$$ This is certainly true if $A+C\ge0$, so we may assume $A+C\lt0$. Then (3) is equivalent to $$(A-C)^2+4B^2\gt(A+C)^2\tag4$$ which is equivalent to $$B^2\gt AC\tag5$$ Now from $A+4B+5C\gt0$ we get $$4B\gt-A-5C\ge2\sqrt{5AC}\tag6$$ where we have used the inequality of the arithmetic and geometric means. But from (6) we get $$B^2\gt(5/4)AC\gt AC\tag7$$ which is (5), and we're done. 
A: With the hint by Cocopuffs, $A + 4B + 5C = \mathrm{tr} (\begin{pmatrix} A & B \\ B & C \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix})$. You only need to say $\begin{pmatrix} A & B \\ B & C \end{pmatrix}$ having two negative eigenvalues is not possible. 
Fact: Let $X, Y$ be $n\times n$ positive semidefinite matrices, then $tr XY\ge 0$.
