Show that $\sqrt{ESS} \leq \sqrt{ESS_{A}}+\sqrt{ESS_{\bar{A}}}$ where ESS=Explained sum of squares Suppose we have a dependent variable $Y$ with mean zero and set of regressors which we divide into two sets, $A$ and $\bar{A}$. Let $ESS$ denote the explained sum of squares (ESS) from regressing $Y$ on all regressors. Let $ESS_A$ denote the ESS from regressing $Y$ on the set of regressors $A$,. and let $ESS_{\bar{A}}$ denote the ESS from regressing $Y$ against regressors $\bar{A}$. 
I want to show that:
$$ \sqrt{ESS} \leq \sqrt{ESS_{A}}+\sqrt{ESS_{\bar{A}}}. $$
Clearly, $ESS\geq ESS_{A}$ and $ESS\geq ESS_{\bar{A}}$, but I do not know how to prove the above, although I am relatively sure the statement holds. Any ideas are highly appreciated. 
 A: Let us interpret this situation geometrically.  When $Y$ is fit to both $A$ and $\bar A$, it is projected to the vector $\hat Y$ in the space spanned by $A$ and $\bar A$.  Because this space contains the spans of $A$ and $\bar A$ separately, the separate fits of $Y$ can be obtained by projecting $\hat Y$ into those subspaces, giving $\hat Y_A$ and $\hat Y_{\bar A}$.  Therefore, we can understand the situation in full generality by drawing a picture of the space spanned by $\hat Y_A$ and $\hat Y_{\bar A}$, because everything happens within this two-dimensional subspace.

Assuming the constant terms have also been accounted for (which means we have already taken the means out of all vectors), the "explained sums of squares" are literally the squared lengths of the vectors:
$$ESS = |\hat Y|^2,\ ESS_A = |\hat Y_A|^2,\ ESS_{\bar A} = |\hat Y_{\bar A}|^2.$$
It should now be geometrically obvious when $\hat Y_A$ and $\hat Y_{\bar A}$ are at obtuse angles--that is, negatively correlated--that $ESS$ can (greatly) exceed the sum of the other two terms.  It is also clear--as in the figure--that $|\hat Y|$ can exceed the sum of the other two lengths, which can be made arbitrarily small by opening up the angle and placing $\hat Y$ along its bisector.
To create a numerical counterexample all we have to do it specify a particular angle $\alpha$ between  $\hat Y_A$ and $\hat Y_{\bar A}$.  Let's make these two vectors have equal lengths for simplicity.  Then we will inject them into a higher-dimensional space in order to create a dataset: this is readily done by choosing any two vectors in $\mathbb{R}^n$ which are both orthogonal and orthogonal to $(1,1,\ldots, 1)$ as a basis for the $\{\hat Y_A, \hat Y_{\bar A}\}$ plane and taking suitable linear combinations of them.  For a little more realism, in the following code I also create a third orthogonal vector for the residuals (the component of $Y$ orthogonal to the space spanned by $A$ and $\bar A$).
a <- 2/3 * pi                                    # Angle between the vectors
n <- c(-1,1); x <- as.matrix(expand.grid(n,n,n)) # Three orthogonal vectors
g <- function(t, i=0) x %*% c(cos(t), sin(t), i) # Special linear combinations
u <- g(0)                                        # One explanatory variable
v <- g(a)                                        # Another variable
y <- g(a/2, 1)                                   # A dependent variable, with error

(You can check that the mean of $y$ is $0$, as assumed in the question, although that does not matter.)
Let's compare the explained sums of squares:
e <- function(f) sum(resid(lm(f))^2)             # Sum of squares of residuals
tss <- e(y~1)                                    # Total sum of squares
c(ESS=tss-e(y~u+v), ESS.u=tss-e(y~u), ESS.v=tss-e(y~v))

The output is
ESS ESS.u ESS.v 
  8     2     2 

Clearly it is not the case that $8 \lt 2 + 2$. 

In response to an edit to the original question, by choosing an angle between $2\pi/3$ and $\pi$ we can even arrange that the square root of ESS exceeds the sum of the square roots of the other two terms.
