Left multiplication with the right Perron eigenvector has a bigger norm? Framework
Let $D$ be a strongly connected, directed graph with adjacency matrix $A\in\{0,1\}^{n\times n}$. Let $\lambda$ be the corresponding Perron eigenvalue, whose corresponding left eigenvector is $v$, and whose right eigenvector is $w$. Moreover, let $v$ and $w$ be such that $\|v\|=\|w\|=1$. 
Observation
In numerical experiments, I consistently find that 
$$ \|v^\top A\|^2 = v^\top A A^\top v \geq w^\top A A^\top w = \|w^\top A\|^2 = \lambda^2 $$
and 
$$ v^\top w <= 1,$$
with equality if and only if $A$ is symmetric. I have tested some 500.000 randomly generated directed graphs, so I'm starting to suspect that there might be some truth to the inequalities above. 
Question
So far, I have not really been able to come up with an argument to support the claim that the above inequalities hold for any (strongly connected) digraph. I was wondering whether anyone would be able to point me in the right direction.  
 A: There seems to be a typo in your question. If you mean $w^TA=\lambda w^T$, then $w$ is a left eigenvector rather than a right eigenvector of $A$.
At any rate, your observation has little to do with adjacency matrices of strongly connected graphs. Suppose $A$ is any real square matrix with a real unit right eigenvector $v$ and a real unit left eigenvector $w$ for the same eigenvalue $\lambda\in\mathbb R$, so that $Av=\lambda v$ and $w^T A=\lambda w^T$. Then
$$
\|w^T A\|=|\lambda|=|v^T(Av)|=|(v^T A)v|\le\|v^T A\|\|v\|=\|v^T A\|.\tag{1}
$$
Similarly,
$$
\|Av\|=|\lambda|=|(w^T A)w|=|w^T(Aw)|\le\|w\|\|Aw\|=\|Aw\|.\tag{2}
$$
By Cauchy-Schwarz inequality, equality holds in $(1)$ if and only if $v^T A$ is a scalar multiple of $v^T$, i.e. if and only if $v$ is also a left eigenvector of $A$. Similarly, equality holds in $(2)$ if and only if $w$ is also a right eigenvector of $A$. This occurs when $A$ is symmetric, but this may also happen otherwise. E.g.
$$
A=\pmatrix{0&1&0\\ 0&0&1\\ 1&0&0}
$$
isn't symmetric, but $v=w=\frac{1}{\sqrt{3}}(1,1,1)^T$ is both a left and a right Perron vector of $A$ and $\|v^TA\|=\|w^TA\|=\|Av\|=\|Aw\|=1$.
