Q:

Compare quadratic term and norm $x^TPx$ and $\|P\|\|x\|^2$ with $P\succ 0$

How to compare the following:   "$x^TPx$ and $\|P\|\|x\|^2$" with $P\succ 0$?  
I mean 
$$x^TPx \geq \|P\|\|x\|^2 \text{ or } x^TPx \leq \|P\|\|x\|^2, \forall x\ne 0, P\succ 0$$    

I know that $$x^TPx = \langle P,X\rangle = \operatorname{tr}(PX)=\operatorname{tr}(Pxx^T)$$
But I still cannot figure out how to go to the step to compare both of them.

A:

When $P$ is positive it is in particular symmetric, whence diagonalizable in an orthonormal basis and having non-negative eigenvalues $\lambda_1\geq \lambda_2 \geq ... 0 $:
$$ P = \sum_i \lambda_i e_i e_i^T $$
So you have (using 2-norm)
$$x^T Px = \sum_i \lambda_i (x^T e_i)^2 \leq \lambda_1 \sum_i (x^T e_i)^2 = \lambda_1 |x|^2 \leq \|P \| \; |x|^2$$ 

