A neat proof of equality involving projection matrices that is reminiscent of Cauchy-Schwarz Trying to prove: 
$$
\|X^\top(I - H_0)Y\|^2= \|(H-H_0)Y\| \|(I - H_0)X\| \tag{1}
$$ 
Here's where this expression comes from. 
In a linear regression 
$$
Y = X\beta + \varepsilon,
$$
I define two (standard) projection matrices. The projection matrix into subspace spanned by columns of the design matrix $X$:
$$
H  := X(X^\top X)^{-1} X^\top,
$$
and projection into the one dimensional subspace spanned by vector $(1,\ldots, 1)$:
$$
H_0 := \frac{1}{n}\mathbf{1} \mathbf{1}^\top.
$$
(Note, one of columns of $X$,  by convention, is a vector $(1,\ldots, 1)$, so we must have $HH_0 = H_0$).
According to my calculations (based on a result from linear regression, where R-sqaure equals square of sample correlation coefficient, $R^2 = r_{xy}^2$, please, see below), it must be true that:
$$
\|X^\top(I - H_0)Y\|^2= \|(H-H_0)Y\| \|(I - H_0)X\| \tag{1}
$$ 
which I find a bit weird, it reminds me of Cauchy-Schwarz, but I couldn't decipher it in this way.
My question: 
Is there an easy way (e.g., geometric, or inner product interpretation) to see why $(1)$ must be true? 
The details are below. 
Note
Here I've asked this question on Cross Validated, now I think the question might be on the linear algebra side, so I've decided to post it here as well and cross-reference. If I make progress I'll leave out one question only to avoid duplicates and update it with an answer.


Details:
With the above projection matrices, $H, H_0$ define the standard quantities associated with a linear regression:
\begin{align}
S_{YY}  &:= \sum_{i=1}^n(y_i - \bar{y})^2 = \|(I - H_0)Y\|^2\,, \\
S_{XX}  &:= \sum_{i=1}^n(x_i - \bar{x})^2 = \|(I - H_0)X\|^2\,, \\
S_{XY}  &:= \sum_{i=1}^n(x_i - \bar{x})(y_i - \bar{y}) = \|X^\top(I - H_0)Y\|^2  = \|Y^\top(I - H_0)X\|^2\,,\\
R_{SS} &:= \sum_{i=1}^n(y_i - \hat{y}_i)^2  = \|(I-H)Y\|^2\,,\\
SS_{reg} &:= \sum_{i=1}^n(\hat{y}_i - \bar{\hat{y}_i})^2  =  \sum_{i=1}^n(\hat{y}_i - \bar{{y}_i})^2   = \|(H-H_0)Y\|^2\,.
\end{align}
Now, on the one hand, 
$$
R^2:= \frac{\sum_{i=1}^n(\hat{y}_i - \bar{\hat{y}_i})^2 }{\sum_{i=1}^n(y_i - \bar{y})^2} = \frac{SS_{reg} }{S_{YY}} = \frac{\|(H-H_0)Y\|^2}{\|(I - H_0)Y\|^2},
$$
and on the other hand 
$$
r^2_{xy}:= \frac{(\sum_{i=1}^n(x_i -\bar{x})(y_i -\bar{y}))^2}{\sum_{i=1}^n(x_i -\bar{x})^2\sum_{i=1}^n(y_i -\bar{y})^2} = \frac{S_{XY}^2}{S_{XX}S_{YY}} = \frac{ \|X^\top(I - H_0)Y\|^4}{\|(I - H_0)X\|^2\, \|(I - H_0)Y\|^2}.
$$
It is a well known fact that the square of sample correlation coefficient and R squared are equal, $r_{xy}^2 = R^2$, which yields that
$$
\frac{ \|X^\top(I - H_0)Y\|^4}{\|(I - H_0)X\|^2\, \|(I - H_0)Y\|^2} =  \frac{\|(H-H_0)Y\|^2}{\|(I - H_0)Y\|^2}.
$$ 
Or equivalently 
$$
\|X^\top(I - H_0)Y\|^2= \|(H-H_0)Y\| \|(I - H_0)X\|. 
$$ 
The last expression looks weird, it remindes me of Cauchy-Schwarz, but I was not able to "decipher" it in this way, is there an easy why to see why $(1)$ must be true? 
Would appreaciate any help. 
 A: Define the special matrices
$$\eqalign{
C &= (I - H_0) \qquad&\big({\rm Centering\,Matrix}\big) \\
H &= X(X^TX)^{-1}X^T \qquad&\big({\rm Hat\,Matrix}\big) \\
}$$
Both are orthoprojectors 
$$\eqalign{
C^2 &= C = C^T \\
H^2 &= H = H^T \\
}$$
and $H$ has special relationships with $X$ and $H_0$
$$\eqalign{
HX &= X &\implies X^T = X^TH \\
HH_0 &= H_0  &\implies H_0 = H_0H \\
}$$
Expand their mutual product to see that they commute with each other.
$$\eqalign{
HC &= (HI-HH_0) = (H-H_0) \\
CH &= (IH-H_0H) = (H-H_0) \\
}$$
Now expand the key quantity of the current problem, and evaluate its sub-multiplicative norm
$$\eqalign{
Q = X^T(I-H_0)Y &= X^TCY \\
 &= (X^TH)CCY \\
 &=  X^TCHCY \\\
\|Q\|=\|X^TC\cdot HCY\| &\le \|X^TC\|\cdot\|HCY\| \\
}$$
So my conclusion is the following inequality
$$\eqalign{
\|X^T(I-H_0)Y\| \;\le\; \|(I-H_0)X\|\cdot\|(H-H_0)Y\| \\
}$$
Similarly, lots of other inequalities can be derived:
$$\eqalign{
\|Q\|
 &\le \|(I-H_0)X\|\cdot\|Y\| \\
\|Q\|
 &\le \|X\|\cdot\|(I-H_0)Y\| \\
\|Q\|
 &\le \|(H-H_0)X\|\cdot\|Y\| \\
\|Q\|
 &\le \|X\|\cdot\|(H-H_0)Y\| \\
\|Q\|
 &\le \|(I-H_0)\|\cdot\|X\|\cdot\|Y\| \\
\|Q\|
 &\le \|(H-H_0)X\|\cdot\|(H-H_0)Y\| \\
}$$
