The inequality is true (and perhaps not "implausible").
Taking logarithms and then derivatives with respect to $t$ one can see that the unique maximum of
$L(t)\stackrel{\text{def}}{=}\left(\frac{t}{d}\right)^{d/2}\left(\frac{d+a}{t+a}\right)^{(d+a)/2} t^{1+\epsilon}C^{-1}$
is at $t_{\text{max}}=\frac{a(d+2+2\epsilon)}{a-2-2\epsilon}$. Assuming $a\ge 3+2\epsilon$ then $\frac{t_{\text{max}}}{d}\le(3+2\epsilon)^2$; picking $C> (3+2\epsilon)^2$ makes $t_{\text{max}}$ fall outside the allowed range $t>Cd$. So one can substitute $t=Cd$ and restate the inequality $L\stackrel{?}{\le}1$ as
$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1\tag{1}$
CASE 1. It's possible to rewrite the term $\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}$ so as to make use of the trivial inequality $(1-\frac{1}{1+z})^{z}<1/\sqrt{3}\space$ if $z\ge1/2$, which implies $\space (1-\frac{y}{x+y})^x<\sqrt{3}^{-y}\space$ if $\space x\ge y/2>0$:
$\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}=\left(1-\frac{(C-1)d/2}{Cd/2+a/2}\right)^{d/2+a/2}$
$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot (Cd)^{1+\epsilon}C^{-1}<C^{d/2}\cdot \sqrt{3}^{-(C-1)d/2}\cdot C^\epsilon d^{1+\epsilon}$
and the condition $\space x\ge y/2>0$ translates into $\space a\ge (C-3)d/2\space$ and $\space C>1$.
Since $\frac{C}{\sqrt{3}^{C-1}}\rightarrow 0\space$ as $\space C\rightarrow\infty$, it's clear that given $\epsilon$, if $C$ is large enough $L<1$ for all $d$'s, provided $a\ge (C-3)d/2\space$.
CASE 2. With $C$ determined by the previous case, assume now $\space K\log(d)<a< (C-3)d/2$, for some $K$. Rewrite (1) as
$L=C^{-a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$
It's possible to rewrite the term $\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}$ so as to make use of the trivial inequality $\space (1+\frac{1}{z})^z<e\space$ if $\space z>0$, which implies $\space (1+\frac{x}{y})^y<e^x\space$ if $\space x>0$ and $y>0$:
$\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}=\left(\frac{d+a}{d+a/C}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\\ =\left(1+\frac{(1-1/C)a/2}{d/2+a/(2C)}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}<e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}$
So now it suffices to prove
$C^{-a/2}\cdot e^{(1-1/C)a/2}\left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$
or
$e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{-a/(2C)} \cdot d^{1+\epsilon}C^{\epsilon}\stackrel{?}{\le}1$
Raise to the power of $2C/a$:
$e^{C-1} \cdot \left(\frac{d+a}{Cd+a}\right)^{C} \cdot \frac{d+a/C}{d+a} \cdot d^{(1+\epsilon)2C/a}\cdot C^{\epsilon 2C/a}\stackrel{?}{\le}1$
Since $a< (C-3)d/2\space$ implies that $\frac{d+a}{Cd+a}<1/3$, and since $\frac{d+a/C}{d+a}<1$ and $a\ge1$, it's sufficient to prove
$\left(\frac{e}{3}C^{2\epsilon}\right)^C \cdot e^{-1} \cdot d^{(1+\epsilon)2C/a}\stackrel{?}{\le}1$
Now $\epsilon$ can made small enough that $\frac{e}{3}C^{2\epsilon}<1$. (Reducing $\epsilon$ does not invalidate CASE 1.)
Last, $a>K\log(d) \implies d^{(1+\epsilon)2C/a}<e^{(1+\epsilon)2C/K}$. All that's left then is to pick $K$ such that $e^{(1+\epsilon)2/K}\cdot \frac{e}{3}C^{2\epsilon}<1$.