Prove $(g^TBg)(g^TB^{-1}g)\ge(g^Tg)^2$ if $B$ is positive definite I'm trying to prove this inequality
$$(g^TBg)(g^TB^{-1}g)\ge(g^Tg)^2,$$
when $B$ is a positive definite matrix. I got a hint that Cauchy-Schwarz inequality can be used.
I tried this way. Define $a=B^Tg$ and $b=B^{-1}g$ then the original inequality becomes
$$(a^Tg)(g^Tb)\ge(g^Tg)^2,$$
when $a^Tb=g^Tg$ and $B$ is positive definite. I found it looks like the form of Cauchy inequality but not exactly, and I have no clue about how to use the positive definiteness of $B$. Any hint help me continue ? 
Update: I guess I made a mistake about the definition of $a$. A transpose is required. 
 A: Because $B$ is positive definite, it has an inverse $B^{-1}$ square root $B^{1/2}$. (Can be obtained, for example, by considering the eigendecomposition of $B$.) Note that $B^{1/2}$ is also positive definite with inverse $B^{-1/2}$.
I will prove the more general statement $(g^\top h)^2 \le (g^\top B g)(h^\top B^{-1} h)$, from which your inequality can be derived by setting $h=g$. Note that this inequality is a special case of dual norms in the case of Mahalanobis distance.
\begin{align}
(g^\top B g)(h^\top B^{-1} h)
&= \|B^{1/2} g\|^2 \|B^{-1/2} h\|^2\\
&\ge ((B^{1/2} g)^\top (B^{-1/2} h))^2 & \text{Cauchy-Schwarz}\\
&= (g^\top h)^2.
\end{align}
A: For a general inner product the Cauchy inequality states that
$$\forall a,b,\, \langle a,b\rangle^2\leq \langle a,a\rangle\langle b,b\rangle$$
Now because $B$ is definite positive $\langle a,b\rangle=a^TBb$ defines an inner product and applying the Cauchy inequality to $a=g$ and $b=B^{-1}g$ one gets using that $B^{-1}$ is symmetric (in fact definite positive)
$$\left(g^TBB^{-1}g\right)^2\leq \left(g^TBg\right)\left( g^TB^{-1}BB^{-1}g\right)$$
And this is the desired inequality.
We have setting $a=g$ and $b=B^{-1}h$ the more general inequality
$$\left(g^Th\right)^2\leq \left(g^TBg\right)\left(h^TB^{-1}h\right)$$ 
A: Using the spectral theorem, we may write $B=P^TDP$, where $P$ is orthogonal and $D$ is diagonal with all entries are positive. Let $h=Pg$. Then the problem is equivalent to:
$$(h^TDh)\cdot(h^TD^{-1}h)\geq (h^Th)^2$$
If $h=\sum_{i=1}^nc_ie_i$ and $D=(\lambda_i)_{i=1}^n$, then the LHS is
$$\left(\sum_{i=1}^n\lambda_i\cdot{c_i}^2\right)\cdot\left(\sum_{i=1}^n\frac{1}{\lambda_i}\cdot{c_i}^2\right)$$
and the RHS is
$$\left(\sum_{i=1}^n{c_i}^2\right)^2$$
Can you now see this as a special case of Hölder's Inequality?
