Inequality with 2-norm of matrix-vector product With the matrix $A \in \mathbb{R}^{m \times n}$, and vector $x \in \mathbb{R}^n$, I'm trying to figure out this inequality:
$$\|Ax\|^2 \leq L\|x\|^2$$ with $L \in \mathbb{R}$, where $L$ is proposed to be the sum of the squares of the entries in the matrix.
I tried expanding to summation notation, but I'm having trouble showing the inequality:
$$
\sum_{i=1}^m \left( \sum_{j=1}^n a_{ij}x_j \right)^2 \leq \left( \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2 \right) \sum_{k=1}^n x_k^2
$$
Is there some inequality that can be used with the square on the left side, since I'm assuming expanding the square will not be helpful?

This is  the Cauchy–Bunyakovsky–Schwarz inequality. For any given $i$, your equation becomes
$$\left(\sum_{j=1}^n a_{ij}x_j\right)^2\le\sum_{j=1}^n a_{ij}^2\sum_{k=1}^n x_k^2$$
Following wikipedia, we can start with a quadratic polynomial
$$0\le(a_{i1}y+x_1)^2+(a_{i2}y+x_2)^2+...+(a_{in}y+x_n)^2=y^2\sum_{j=1}^n a_{ij}^2+2y\left(\sum_{j=1}^n a_{ij}x_j\right)+\sum_{j=1}^n x_j^2$$ 
The polynomial is always grater or equal to $0$, which means it cannot have two real solutions. That means that the discriminant is less then $0$
$$4\left(\sum_{j=1}^n a_{ij}x_j\right)^2-4\sum_{j=1}^n a_{ij}^2\sum_{k=1}^n x_k^2\le 0$$
