Here is how to remove the assumption that $p-2+\delta\ge0$.
Let
\begin{equation*}
s:=p-1+\delta.
\end{equation*}
The conditions $p>1$ and $\delta>0$ imply $s\ge0$. No other conditions on $p$ and $s$ will be used or needed in what follows.
The inequality in question will follow from the inequality
\begin{equation*}
L\le R(s)
\end{equation*}
for
\begin{equation*}
s\in[0,\infty),\ p\in[1,\infty),\ 0<a\le1\le b,
\end{equation*}
where
\begin{equation*}
L:=2^{1-p} (a+b)^p,\\
R(s):=R(s,p):=a^p \Big(1+\ln\frac{a+b}2\Big)^{-s}+b^p \Big(\frac{1+\ln b}{1+\ln\frac{a+b}2}\Big)^s.
\end{equation*}
Clearly, $R(s)$ is convex in $s$.
Lemma 1: $R'(0)\ge0$, and hence $R(s)$ is increasing in $s$.
By Lemma 1 and the power-means inequality,
\begin{equation*}
R(s)\ge R(0)=a^p + b^p\ge 2^{1-p} (a+b)^p=L,
\end{equation*}
as desired.
It remains to provide
Proof of Lemma 1: Let
\begin{equation*}
R_1(p,a,b):=R'(0)/a^p \\
=
\Big(\frac{b}{a}\Big)^p
\ln\frac{1+\ln b}{1+\ln\frac{a+b}{2}}
-\ln\Big(1+\ln\frac{a+b}{2}\Big),
\tag{0}\label{0}
\end{equation*}
which is clearly nondecreasing in $p$. So,
\begin{equation*}
aR_1(p,a,b)\ge aR_1(1,a,b) \\
=b \ln (1+\ln b)
-(a+b) \ln\Big(1+\ln\frac{a+b}{2}\Big),
\end{equation*}
which latter is clearly decreasing in $a$. So,
\begin{equation*}
aR_1(p,a,b)\ge aR_1(1,a,b)\ge R_1(1,1,b)=bg(b),
\tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
g(b):=\ln (1+\ln b) -\frac{1+b}b\,\ln\Big(1+\ln\frac{1+b}2\Big).
\end{equation*}
So, in view of \eqref{0}, it remains to prove that
\begin{equation*}
g(b)\overset{\text{(?)}}\ge0 \tag{2}\label{2}
\end{equation*}
(for real $b>1$).
To begin the proof of \eqref{2}, let
\begin{equation*}
g_1(b):=g'(b)b^2 \\
= b \Big(\frac{1}{1+\ln b}-\frac{1}{1+\ln\frac{1+b}2}\Big)
+\ln\Big(1+\ln\frac{1+b}{2}\Big). \tag{3}\label{3}
\end{equation*}
Now use the substitution $b=2e^t-1$, so that $t>0$ and $\ln\frac{1+b}{2}=t$. Let
\begin{equation*}
\begin{aligned}
G_2(t)&:=\frac{dg_1(2e^t-1)}{dt}\,
\frac{(1+t)^2 (1+\ln (2 e^t-1))^2}{(2 e^t-1)t} \\
& =\frac{2 (t+e^t (1+t^2)) \ln (2e^t-1)}{(2 e^t-1) t}-1-\ln ^2(2 e^t-1).
\end{aligned}
\tag{4}\label{4}
\end{equation*}
Next, note that
\begin{equation*}
h(t):=t^3+t^2+t+2 e^t \left(t^2+1\right)-1>0
\end{equation*}
(for $t>0$) and let
\begin{equation*}
\begin{aligned}
G_3(t)&:=G_2'(t)\frac{t^2(2 e^t-1)^2}{2e^t h(t)} \\
& =\frac{2 t (t + e^t (1 + t^2))}{h(t)}- \ln(2e^t-1),
\end{aligned}
\tag{5}\label{5}
\end{equation*}
\begin{equation*}
G_4(t):=G'_3(t)\frac{(2e^t-1)h(t)^2}{2t} \\
=2 - t + t^3 + 2 e^{2 t} (1 - 7 t - 4 t^2 - 2 t^3 - t^4 + t^5) -
e^t (5 - 8 t + 6 t^3 + 3 t^4 + 2 t^5),
\tag{5.5}\label{5.5}
\end{equation*}
\begin{equation*}
G_5:=G_4',\ G_6:=G_5',\ G_7:=G_6', \tag{6}\label{6}
\end{equation*}
\begin{equation*}
G_8(t):=G_7'(t)\frac{e^{-t}}{P(t)}
=16 e^t \frac{Q(t)}{P(t)}-1,
\tag{7}\label{7}
\end{equation*}
where
\begin{equation*}
P(t):=189 + 736 t + 768 t^2 + 294 t^3 + 43 t^4 + 2 t^5,\ \\
Q(t):=-65 - 91 t - 8 t^2 + 40 t^3 + 18 t^4 + 2 t^5.
\end{equation*}
Next,
\begin{equation*}
G'_8(t)=16 e^t \frac{S(t)}{P(t)^2}>0,
\end{equation*}
where
\begin{equation*}
S(t):=18356 + 31777 t + 25602 t^2 + 49850 t^3 + 84244 t^4 + 72903 t^5 +
34758 t^6 + 9548 t^7 + 1492 t^8 + 122 t^9 + 4 t^{10}.
\end{equation*}
So, $G_8(t)$ is increasing (in $t>0$). Also, $G_8(0)<0$ and $G_8(\infty-):=\lim_{t\to\infty}G_8(t)=\infty>0$. So, $G_8$ is $-+$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G_8<0$ on $(0,c)$ and $G_8>0$ on $(c,\infty)$.
So, by \eqref{7}, $G_7$ is down-up (on $(0,\infty)$) -- that is, for some real $c>0$ we have that $G_7$ is decreasing on $(0,c)$ and increasing on $(c,\infty)$.
Also, $G_7(0)<0$ and $G_7(\infty-)=\infty>0$. So, $G_7$ is $-+$.
So, by \eqref{6}, $G_6$ is down-up.
Also, $G_6(0)<0$ and $G_6(\infty-)=\infty>0$. So, $G_6$ is $-+$.
So, by \eqref{6}, $G_5$ is down-up.
Also, $G_5(0)<0$ and $G_5(\infty-)=\infty>0$. So, $G_5$ is $-+$.
So, by \eqref{6}, $G_4$ is down-up.
Also, $G_4(0)<0$ and $G_4(\infty-)=\infty>0$. So, $G_4$ is $-+$.
So, by \eqref{5.5}, $G_3$ is down-up.
Also, $G_3(0)=0$ and $G_3(\infty-)=-\ln2<0$. So, $G_3<0$.
So, by \eqref{5}, $G_2$ is decreasing.
Also, $G_2(0+):=\lim_{t\downarrow0}G_2(t)=3>0$ and $G_2(\infty-)=-\infty<0$. So, $G_2$ is $+-$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G_2>0$ on $(0,c)$ and $G_2<0$ on $(c,\infty)$.
So, by \eqref{4}, $g_1$ is up-down -- that is,
for some real $c>1$ we have that $g_1$ is increasing on $(1,c)$ and decreasing on $(c,\infty)$.
Also, $g_1(1)=0$ and $g_1(\infty-)=-\infty<0$. So, $g_1$ is $+-$ (on $(1,\infty)$).
So, by \eqref{3}, $g$ is up-down.
Also, $g(1+)=0$ and $g(\infty-)=0$. Thus, \eqref{2} follows. $\quad\Box$
Best Answer
If $y^2$ is defined by (1) and $$\epsilon = \frac{168}{97},x= \frac{1168561}{2916},z= \frac{121}{8281},\Delta = \frac{1}{1000},K= \frac{10201}{9604},$$ then the difference between the right-hand side of inequality (2) and its left-hand side is $-2.5248\ldots<0$, so that (2) fails to hold.
(What might be interesting, though, is that, for $y^2,\epsilon,x,z,\Delta,K$ as above, the values of the right-hand side and left-hand side are relatively close to each other: $2635.1\ldots$ and $2637.6\ldots$.)
Also, if $y^2$ is defined by (1), $y>0$, $V=y^{-1}\left(2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right)$ and $$K= \frac{602}{61},x= \frac{7936}{57},\delta = \frac{34618}{41},\epsilon = \frac{6}{73},\Delta = \frac{37534}{29},$$ then the ratio of the left-hand side of inequality $$\epsilon V \le K^{-\frac 12} \tag{*}$$ to its right-hand side is $1.2496\ldots>1$, so that (*) fails to hold, too.