Vtyjkyuk

Suppose $A=(a_ij)$ is a Hermitian $ntimes n$ matrix such that the symmetric matrix $vert Arvert=(lvert a_ijrvert)$ has a largest eigenvalue $lambda_1(lvert Arvert)=1$, and all other eigenvalues satisfy $$frac12ge lambda_2(lvert Arvert)gedotsge lambda_n(vert Arvert)ge-frac12.$$

Let $x$ be the normalised eigenvector of $A$ corresponding to eigenvalue $lambda_1(A)$. Then
$$lambda_1(A)=langle x,Axrangle = lvertlangle x,Axranglervert le langle lvert xrvert,lvert Arvertlvert xrvertranglele lambda_1(lvert Arvert),$$
but what can be said about $lambda_2(A)$? For example is it true that for sufficiently small $epsilon>0$ there exists a small $delta=delta(epsilon)>0$ such that
$$lvert lambda_1(A)-1rvert<deltaquadimpliesquad lambda_2(A)lefrac12+epsilon?$$
If yes, how can $delta(epsilon)$ be chosen?

The answer to your question is yes.

First of all, consider the case $lambda_1(A) = 1$.
Note that $A_ij = 0$ if $x_i ne 0$ and $x_j = 0$: otherwise you could increase $langle y, |A| y rangle/langle y, yrangle$ by taking $y = |x| + t e_j$ for some small $t$ (the numerator is $1 + c t + O(t^2)$ for some $c > 0$, the denominator $1 + O(t^2)$).
Thus (reordering the rows and columns if necessary) if some $x_i = 0$ the matrix $A$ has block form $pmatrixB & 0cr 0 & Ccr$, and its eigenvalues are the union of the eigenvalues of $B$ and of $C$, and similarly for $|A|$ with $|B|$ and $|C|$; $1$ is an eigenvalue of $B$ and of $|B|$ while all eigenvalues of $|C|$ are at most $1/2$. Then all eigenvalues of $C$ have the same bound. Thus we can restrict our attention to $B$, i.e. we may assume all $x_i ne 0$.

Let $U$ be a diagonal matrix with diagonal entries $u_ii = x_i/|x_i|$.
We have $langle x, A x rangle = langle |x|, |A| |x| rangle = 1$ which
implies all the terms $overlinex_i A_ij x_j ge 0$, i.e. $|x_i| overlineu_ii A_ij u_jj |x_j| ge 0$, so $overlineu_ii A_ij u_jj ge 0$. This says $|A| = U^* A U$. But then $|A|$ and $A$ have the same eigenvalues, so $|lambda_2(A)| le 1/2$.

Finally, note that the set of Hermitian matrices satisfying your condition is compact.
If the answer were no, then there would be $epsilon > 0$ and a sequence of such matrices $A_k$ with
$lambda_1(A_k) to 1$ and $lambda_2(A_k) > 1/2 + epsilon$. Taking a limit point, we get $A$ satisfying the condition with $lambda_1(A) = 1$ and $lambda_2(A) ge 1/2 + epsilon$, which as we have seen is impossible.