Quadratic form $x^T A x $ is less than $||A|| x^T x$? Does the following hold?
$$x^T A x \leq ||A|| x^T x$$
$A$ is a symmetric positive-definite real-valued matrix, and $x$ is a real-valued vector. Norm is Euclidean.
 A: First, since $A$ is SPD, write 
$$
A = Q^t D Q
$$
for some orthogonal $Q$ and diagonal $D$ with all diagonal entries $d_i > 0$. Write each $d_i = c_i^2$, (Symmetric matrices are diagonalizable, by Sylvester's law of inertia). 
Second, consider the case where $x^t x = 1$, because the statement is true if and only if it's true for this special case. So now we're asking:
For all $x$ with $x^t x = 1$, and all orthogonal $Q$ and diagonal $D$, is it true that 
$$
x^t Q^t D Q x \le \| D \| ?
$$
Letting $y = Qx$, we see that as $x$ ranges over the unit sphere, $y$ does so as well (because $Q$ is length preserving and invertible). So we can equally well ask:
For all $y$ with $y^t y = 1$, and all  diagonal $D$, is it true that 
$$
y^t D  y \le \| D \| ?
$$
The left hand side is 
$$
\sum_i d_i y_i^2
=\sum_i c_i^2 y_i^2
=\sum_i (c_i y_i)^2
$$
That is to say, it gives the squared lengths of all vectors in an ellipsoid that's the result of stretching a sphere by factor $c_i$ along the $i$th axis. The largest possible value here is the length of the largest semi-axis of the ellipsoid, i.e., $\max_i c_i^2 = \max_i d_i$. 
The right hand side is 
$$
\sqrt{\sum_i d_i^2} 
 \ge \sqrt{ \max_i d_i^2 } = \max_i d_i
$$
from which the result follows. 
