$A+cI$, positive? $c>\sum\limits_{i,j} |a_{ij}|$, proof verification. 
Let $A$ be a real symmetric matrix. Show that there exists a real number $c$ so that $A+cI$ is positive.
SOLUTION:
$$X^tAX=\sum_{i,j} a_{ij} x_ix_j\geqslant\sum_{i,j} -|a_{ij}||x_k|^2$$
for some $k$, so if we choose $c>\sum\limits_{i,j} |a_{ij}|$, then $A+cI$ is positive because:
$X^t(A+cI)X=X^tAX+c(x_1^2+\cdots+x_n^2)$. Solutions Manual for Lang´s Linear Algebra, Rami Sarkarchi.

Questions:
1) Why does the author write $c>\sum\limits_{i,j} |a_{ij}|$? Instead of $c\leqslant\sum\limits_{i,j} |a_{ij}|$?
2) Why write $\sum\limits_{i,j} a_{ij}x_ix_j\geqslant\sum\limits_{i,j} -|a_{ij}||x_k|^2$? Was it not supposed to be $$\sum_{i,j} a_{ij}x_ix_j-\sum_{i,j} |a_{ij}||x_k|^2\geqslant 0\implies\sum_{i,j} a_{ij}x_ix_j\geqslant\sum_{i,j} |a_{ij}||x_k|^2 \text{ ?}$$
Thanks in advance!
 A: The argument shows that as $c$ gets bigger, then so does $X^t(A+cI)X.$ Its goal is to show that if $c$ is big enough, then $X^t(A+cI)X$ will be positive regardless of which vector $X$ is. So a question is, how big does $c$ have to be in order to be big enough to assure that. Therefore one wants to say that if $c>\text{something}$ then that conclusion holds. The question then is: What number should be in the role of "something"? That's why they need to say $c>\text{something.}$
A: For 1): since 
$$X^t\left(   A+cI\right) X\geqslant -\sum_{i,j}\left\lvert a_{i,j}\right\rvert x_k^2 +\sum_{l=1}^ncx_l^2\geqslant      -\sum_{i,j}\left\lvert a_{i,j}\right\rvert x_k^2+c x_k^2 =\left(c-\sum_{i,j}\left\lvert a_{i,j}\right\rvert\right)   x_k^2 ,$$
we need $c-\sum_{i,j}\left\lvert a_{i,j}\right\rvert\gt 0$.
For 2): for any $i,j$, 
$$a_{i,j}x_ix_j\geqslant -\left\lvert  a_{i,j}x_ix_j\right\rvert=  -\left\lvert  a_{i,j}\right\rvert\left\lvert x_i\right\rvert\left\lvert x_j\right\rvert\geqslant  -\left\lvert  a_{i,j}\right\rvert\left\lvert x_k\right\rvert^2,          $$ 
where $k$ is such that $\left\lvert x_k\right\rvert =\max_{1\leqslant l\leqslant n}    \left\lvert x_l\right\rvert$ (in particular, if $X\neq 0$, this guarantees that $ x_k\neq 0$ hence that $A+cI$ is positive definite).    
