Show $x^2 A + x B +C$ is positive semi-definite EDIT: Only $A$ is allowed to be positive semi-definite.
I'm interested in showing that the matrix $f(x)=x^2 A + x B +C$ is positive semi-definite (PSD) for any scalar $x$, and PSD matricex $A$. I use the following definition for a PSD matrix:
\begin{align}
u^T f(x) u &= x^2\, u^T A u + x\, u^T B u + u^T C u \\
&\geq 0
\end{align}
for any vector $u$.
Then I use the quadratic formula to obtain the following condition:
\begin{align}
(u^T B u )^2 - 4(u^T A u)(u^T C u) 
&= u^T B u u^T B u - 4 u^T A u u^T C u \\
&= u^T (B u u^T B - A u u^T C) u \\
&\leq 0
\end{align}
for any vector $u$.
I don't know how to get rid of $u$ and translate this condition to one involving $A$, $B$ and $C$ only.
 A: Partial answer; certainly too long for a comment.
You're looking for conditions on $A,B,C$ that guarantee that for any $u$, we will have 
$$
u^T(Buu^TB - Auu^TC)u = 
\operatorname{Tr}[uu^T(Buu^TB - Auu^TC)] \geq 0
$$
Notably, the map $\langle P,Q \rangle = \operatorname{Tr}(PQ^T)$ defines an inner product over $\Bbb R^{n \times n}$.  If we define $\phi$ to be the linear map given by
$$
\phi(X) = BXB - AXC,
$$
then we're looking for conditions that guarantee $\langle \phi(uu^T), uu^T \rangle \geq 0$ for all choices of unit vector $u$. Equivalently, we're looking for conditions that guarantee $\langle \phi(X), X \rangle \geq 0$ whenever $X$ is a symmetric matrix.
Via the vectorization operator, we see that $\phi$ (over all of $\Bbb R^{n \times n}$) can be represented by the matrix
$$
[\phi] = B^T \otimes B - C^T \otimes A
$$
where $\otimes$ denotes the Kronecker product.  A sufficient (but not necessary, I believe) condition to guarantee that the desired condition holds is that the matrix $[\phi] + [\phi]^T$ is symmetric positive semidefinite.
