Implausible inequality Let $t,d,a \ge 1$, of which $d$ can be unbounded, $a$ can be constrained to be larger than some threshold (which can depend on $d$ in some mild way, say logarithmically). How to find out whether the following true?
$\exists C,\epsilon>0$ constants independent of $d$ s.t. $\forall t>d\cdot C$, it holds that:
$\left(\frac{t}{d}\right)^{d/2}\left(\frac{d+a}{t+a}\right)^{(d+a)/2} \le C\cdot t^{-1-\epsilon}$
What I was thinking to try is to look at the limit of the fraction of the LHS/RHS when $t\rightarrow \infty$, see if it is finite. Well, it is (zero for $1+\epsilon \le a/2$). But that doesn't give a constant $C$ independent of $d$. Ideas would be appreciated.
EDIT: Replaced the condition as suggested by Yaakov Baruch's comment.
 A: The "implausible" inequality is true for any $\epsilon$, for some $C$ and $K$, if $\space a>K\log(d)+K$.
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 $K\ge 3+2\epsilon$ then $\frac{t_{\text{max}}}{d}\le(3+2\epsilon)^2$; so  $t_{\text{max}}$ fall outside the range $t>Cd\space$ if $\space C> (3+2\epsilon)^2$, and one only needs to verify $L\le1$ for $t=Cd$:
$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1\tag{1}$

CASE 1.
$(1-\frac{1}{1+z})^{z}<1/\sqrt{3}\space$ if $\space z\ge1/2 \implies (1-\frac{y}{x+y})^x<\sqrt{3}^{-y}\space$ if $\space x\ge y/2>0$, therefiore:
$\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}=\left(1-\frac{(C-1)d/2}{Cd/2+a/2}\right)^{d/2+a/2}<\sqrt{3}^{-(C-1)d/2}$
$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot (Cd)^{1+\epsilon}C^{-1}<C^{d/2}\cdot \sqrt{3}^{-(C-1)d/2}\cdot C^\epsilon d^{1+\epsilon}$
$\frac{C}{\sqrt{3}^{C-1}}\rightarrow 0\space$ as $\space C\rightarrow\infty$, so given $\epsilon$, the inequality holds for every $d$ if $C$ is large enough, provided $x\ge y/2$, that is  $a\ge (C-3)d/2$.

CASE 2. Now assume $\space K\log(d)+K<a< (C-3)d/2\space$ for some $K$. Rewrite (1) as
$L=C^{-a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$
$\space (1+\frac{1}{z})^z<e\space$ if $\space z>0 \implies (1+\frac{x}{y})^y<e^x\space$ if $\space x,y>0$, therefore:
$\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}=\left(\frac{d+a}{d+a/C}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\\
=\left(1+\frac{(1-1/C)a/2}{d/2+a/(2C)}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}<e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}$
Now it suffices to prove
$C^{-a/2}\cdot \left(e\cdot \frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$
and raising to the power of $2/a$, and rearranging, gives
$e^{1-1/C} \cdot \frac{d+a}{Cd+a} \cdot \left(\frac{d+a/C}{d+a}\right)^{1/C}\cdot C^{\epsilon 2/a} \cdot d^{(2+2\epsilon)/a}\stackrel{?}{\le}1$
Since $a< (C-3)d/2 \implies \frac{d+a}{Cd+a}<1/3$, and since $\frac{d+a/C}{d+a}<1$, it's sufficient to prove
$\frac{e^{1-1/C}}{3}\cdot C^{2\epsilon/a} \cdot d^{(2+2\epsilon)/a}\stackrel{?}{\le}1$
Last, $a>K\log(d)+K \implies d^{(2+2\epsilon)/a}<e^{(2+2\epsilon)/K}$. All that's left then is to choose $K$ such that
$\frac{e}{3}\cdot C^{2\epsilon/K} \cdot e^{(2+2\epsilon)/K}<1$.
