Inequality about modified Jordan form I'm trying to prove an inequality, which in turn I think is equivalent to this inequality,
$$(|\lambda_1|-\epsilon)^2\bar{y}^Ty\le\bar{y}^T\left(\overline{(\lambda_rI_r+\epsilon F_r)_r}^T(\lambda_rI_r+\epsilon F_r)_r\right)y\le(|\lambda_d|+\epsilon)^2\bar{y}^Ty$$ for all $y\in\mathbb{C}^d$, for any $0\le\epsilon\le|\lambda_1|$.
where $(\lambda_rI_r+\epsilon F_r)_r$ is a slightly modified Jordan form of a complex-valued matrix $M$. $\lambda_rI_r+\epsilon F_r$ is the modified $r$-th Jordan block with $I_r$ being identity matrix and $F_r$ being $1$s right above diagonal. $M$ has eigenvalues $\lambda_1,...,\lambda_d$ with order that $|\lambda_1|\le|\lambda_2|\le...\le|\lambda_d|$.
I think the $M$ is not needed here, and more advanced knowledge about Jordan form except the definition neither. I assume the Jordan form of $M$ has $n$ Jordan blocks. Then I wish I could prove that 
$$(|\lambda_1|-\epsilon)^2\bar{y_r}^Ty_r\le\bar{y_r}^T\overline{(\lambda_rI_r+\epsilon F_r)}^T(\lambda_rI_r+\epsilon F_r)y_r\le(|\lambda_d|+\epsilon)^2\bar{y_r}^Ty_r$$
for $r=1,...,n$, where $y_r$ is the corresponding $r$-th block of $y$. Then the middle part can be expanded to
$$\bar{y_r}^T(|\lambda_r|^2I_r+\bar{\lambda_r}\epsilon F_r+\lambda_r\epsilon F_r^T+\epsilon^2F_r^TF_r)y_r.$$
Then the inequality is equivalent to
$$|\lambda_1|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-2|\lambda_1|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i\le|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}+\overline{\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}}+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i\le|\lambda_d|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+2|\lambda_d|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i$$
I can see the close similarity between the tree formulas separated by $\le$, but I am stuck in to show the inequality. Especially how can we handle the conjugate terms ? Or the second inequality which I want to prove is true or not ?
Important Update: I'm so sorry that I missed to impose an essential condition that $0\le\epsilon\le|\lambda_1|$
 A: To prove this inequality $$|\lambda_1|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-2|\lambda_1|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i
\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}+\overline{\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}}+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i
\le
|\lambda_d|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+2|\lambda_d|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i \quad\quad (1)$$
We first prove a useful intermediate result
$$
\begin{align}
-|\lambda_r|\epsilon\left(\sum_{i=2}^{s_r}\bar{y_i}y_i\right)^{1/2}\left(\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}\right)^{1/2}
\le Re\left(\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}\right)
\le
|\lambda_r|\epsilon\left(\sum_{i=2}^{s_r}\bar{y_i}y_i\right)^{1/2}\left(\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}\right)^{1/2}\quad\quad (2)
\end{align}$$
By $Re(z)\le|z|$ we get
$$Re\left(\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}\right)
\le
|\lambda_r|\epsilon\sum_{i=2}^{s_r}|\bar{y_i}y_{i-1}|
\le
|\lambda_r|\epsilon\left(\sum_{i=2}^{s_r}\bar{y_i}y_i\right)^{1/2}\left(\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}\right)^{1/2}
. $$
Similarly we can get the other direction of $(2)$. By this we can easily see that the second inequality of $(1)$ holds.
Now we prove the first inequality of (1).
$$
\begin{align}
|\lambda_1|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-&2|\lambda_1|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i
\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-2|\lambda_r|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i  \\
&= 
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\bar{y_1}{y_1}+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i   \\
&\le 
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_i-\epsilon^2\bar{y_1}{y_1}+\epsilon^2\sum_{i=1}^{s_r}\bar{y_i}y_i   \\
&=
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i   \\
&\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i   \\
&\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}-|\lambda_r|\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_i+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i   \\
&\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i-
2|\lambda_r|\epsilon\left(\sum_{i=2}^{s_r}\overline{y_{i-1}}y_{i-1}\right)^{1/2}\left(\sum_{i=2}^{s_r}\bar{y_i}y_i\right)^{1/2}
+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i   \\
&\le
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+
2Re\left(\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}\right)
+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i   \\
&=
|\lambda_r|^2\sum_{i=1}^{s_r}\bar{y_i}y_i+\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}+\overline{\lambda_r\epsilon\sum_{i=2}^{s_r}\bar{y_i}y_{i-1}}+\epsilon^2\sum_{i=2}^{s_r}\bar{y_i}y_i
\end{align}
$$
The $7$-th line is due to $(a+b)\ge2(ab)^{1/2}$ for nonnegative $a$ and $b$.
