There exists a real number $c$ such that $A+cI$ is positive when $A$ is symmetric Without using the fact that symmetric matrices can be diagonalized: Let $A$ be a real symmetric matrix. Show that there exists a real number $c$ such that $A+cI$ is positive.
That is, if $A=(a_{ij})$, one has to show that there exists real $c$ that makes $\sum_i a_{ii}x_i^2 + 2\sum_{i<j}a_{ij}x_ix_j + c\sum_i x_i^2 > 0$  for any vector $X=(x_1,...,x_n)^T$.
This is an exercise in Lang's Linear Algebra.
Thank you for your suggestions and comments.

You can use the fact
$$
\lambda_{min}x^Tx \leq x^TAx\leq \lambda_{max}x^Tx
$$
(which is not difficult to prove) and suppose that the matrix is already negative definite hence all the eigenvalues are negative. This means $x^TAx<0$ for all non-zero $x$. This allows us to write
$$
\lambda_{min}\|x\|^2 \leq x^TAx\leq \lambda_{max}\|x\|^2 < 0
$$
But, consider the following:
$$
x^T(A+cI)x = x^TAx +cx^Tx \geq (\lambda_{min}+c)\|x\|^2
$$
If we select $c>|\lambda_{min}|$ we obtain a positive definite matrix since for every non-zero $x$, $x^T(A+cI)x > 0$
