$A \in M_n(\mathbb R)$ is symmetric: If one of the entries in its diagonal is positive, prove that it has at least one positive eigenvalue 
Let $A \in M_n(\mathbb R)$ be symmetric. Given that one of the entries in its diagonal is positive, prove that it has at least one positive eigenvalue.

I didn't come to any conclusion. 
Thanks guys.
 A: Let $\ell$ be the linear transformation and $q$ the quadratic form given by $A$: 
$$\ell(x):=Ax,\quad q(x):=x^TAx.$$ 
The assumption that $A$ has a positive diagonal entry means that there is a vector on which $q$ is positive. 
A matrix represents $\ell$ (in an appropriate basis) iff it is of the form $PAP^{-1}$ with $P$ invertible. 
A matrix represents $q$ (in an appropriate basis) iff it is of the form $PAP^T$ with $P$ invertible. 
The theory tells us that, since $A$ is symmetric (*), we can pick a $P$ such that $P^{-1}=P^T$ and $PAP^T$ is a diagonal matrix $D$. In particular, the entries of $D$ are the eigenvalues of $A$. If they were all nonpositive, $q$ would be nonpositive on all vectors. 
(*) See Gerry Myerson's comment to percusse's answer.
A: Suppose the matrix $A$ has no positive eigenvalue, hence a negative (semi)definite matrix $A\preceq 0$ or $x^TAx \leq 0$ for all $x\in\mathbb{R}^n$. 
For simplicity, assume that the positive entry is the $A_{11}$ (or upper left etc.) entry of the matrix. Then, using a vector $x = \begin{bmatrix}1 &0 &\ldots &0\end{bmatrix}^T$, we know that $x^TAx > 0$.
A: Because $A$ is symmetric, it admits real Schur decomposition, i.e. $A= U^t.D.U$. 
Now suppose $A_{1,1}$ is positive, and assume that all eigenvalues are non-positive. Then
we arrive at contradiction:
$$
  A_{1,1} = \sum_{k=1}^n U^t_{1,k} \lambda_k U_{k,1} =  \sum_{k=1}^n \lambda_k \left( U_{k,1} \right)^2 <= 0
$$
