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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

$C^{-a/2}\cdot e^{(1-1/C)a/2}\left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$

and raising to the power of $2/a$, and rearranging, gives

$e^{1-1/C} \cdot \left(\frac{d+a}{Cd+a}\right) \cdot \left(\frac{d+a/C}{d+a}\right)^{1/C}\cdot C^{\epsilon 2/a} \cdot d^{(2+2\epsilon)/a}\stackrel{?}{\le}1$

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 pick $K$ such that

$\frac{e}{3}\cdot C^{2\epsilon/K} \cdot e^{(2+2\epsilon)/K}<1$.

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

$\frac{e}{3}\cdot C^{2\epsilon/K} \cdot e^{(2+2\epsilon)/K}<1$.

The inequality is true (so 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. Use the 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 $\space x\ge y/2>0$:

$\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}=\left(1-\frac{(C-1)d/2}{Cd/2+a/2}\right)^{d/2+a/2}<\sqrt{3}^{-(C-1)d/2}$

$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot (Cd)^{1+\epsilon}C^{-1}<C^{d/2}\cdot \sqrt{3}^{-(C-1)d/2}\cdot C^\epsilon d^{1+\epsilon}$

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 $\epsilon$, the inequality holds for all $d$'s if $C$ is large enough and $a\ge (C-3)d/2\space$.

CASE 2. With $C$ from CASE 1, assume $\space K\log(d)<a< (C-3)d/2\space$ for some $K$. Rewrite (1) as

$L=C^{-a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$

Use the 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 $\space x,y>0$:

$\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}=\left(\frac{d+a}{d+a/C}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\\ =\left(1+\frac{(1-1/C)a/2}{d/2+a/(2C)}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}<e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}$

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 $\frac{d+a/C}{d+a}<1$, it's sufficient to prove

$\left(\frac{e}{3}C^{2\epsilon/a}\right)^C \cdot e^{-1} \cdot d^{(1+\epsilon)2C/a}\stackrel{?}{\le}1$

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 $e^{(1+\epsilon)2/K}\cdot \frac{e}{3}C^{2\epsilon}<1$.

REMARK. If one constrains $a$ slightly differently (that is: $\space a>K\log(d)+K$) then $\frac{e}{3}C^{2\epsilon/a}<1$ for $K$ large enough (without need to reduce $\epsilon$). This proves that for any $\epsilon>0$ there are $C$ and $K$ for which the inequality holds.

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

$C^{-a/2}\cdot e^{(1-1/C)a/2}\left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$

and raising to the power of $2/a$, and rearranging, gives

$e^{1-1/C} \cdot \left(\frac{d+a}{Cd+a}\right) \cdot \left(\frac{d+a/C}{d+a}\right)^{1/C}\cdot C^{\epsilon 2/a} \cdot d^{(2+2\epsilon)/a}\stackrel{?}{\le}1$

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 pick $K$ such that

$\frac{e}{3}\cdot C^{2\epsilon/K} \cdot e^{(2+2\epsilon)/K}<1$.

The inequality is true (so 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. Use the 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 $\space x\ge y/2>0$:

$\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}=\left(1-\frac{(C-1)d/2}{Cd/2+a/2}\right)^{d/2+a/2}<\sqrt{3}^{-(C-1)d/2}$

$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot (Cd)^{1+\epsilon}C^{-1}<C^{d/2}\cdot \sqrt{3}^{-(C-1)d/2}\cdot C^\epsilon d^{1+\epsilon}$

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 $\epsilon$, the inequality holds for all $d$'s if $C$ is large enough and $a\ge (C-3)d/2\space$.

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

$L=C^{-a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$

Use the 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 $\space x,y>0$:

$\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}=\left(\frac{d+a}{d+a/C}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\\ =\left(1+\frac{(1-1/C)a/2}{d/2+a/(2C)}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}<e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}$

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 $\frac{d+a/C}{d+a}<1$, it's sufficient to prove

$\left(\frac{e}{3}C^{2\epsilon/a}\right)^C \cdot e^{-1} \cdot d^{(1+\epsilon)2C/a}\stackrel{?}{\le}1$

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 $e^{(1+\epsilon)2/K}\cdot \frac{e}{3}C^{2\epsilon}<1$.

REMARK. If one constrains $a$ slightly differently (that is: $\space a>K\log(d)+K$) then $\frac{e}{3}C^{2\epsilon/a}<1$ for $K$ large enough (without need to reduce $\epsilon$). This proves that for any $\epsilon>0$ there are $C$ and $K$ for which the inequality holds.

The inequality is true (so 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. Use the 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 $\space x\ge y/2>0$:

$\left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}=\left(1-\frac{(C-1)d/2}{Cd/2+a/2}\right)^{d/2+a/2}<\sqrt{3}^{-(C-1)d/2}$

$L=C^{d/2}\cdot \left(\frac{d+a}{Cd+a}\right)^{d/2+a/2}\cdot (Cd)^{1+\epsilon}C^{-1}<C^{d/2}\cdot \sqrt{3}^{-(C-1)d/2}\cdot C^\epsilon d^{1+\epsilon}$

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 $\epsilon$, the inequality holds for all $d$'s if $C$ is large enough and $a\ge (C-3)d/2\space$.

CASE 2. With $C$ from CASE 1, assume $\space K\log(d)<a< (C-3)d/2\space$ for some $K$. Rewrite (1) as

$L=C^{-a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}\cdot C^\epsilon d^{1+\epsilon}\stackrel{?}{\le}1$

Use the 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 $\space x,y>0$:

$\left(\frac{d+a}{d+a/C}\right)^{d/2+a/2}=\left(\frac{d+a}{d+a/C}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}\\ =\left(1+\frac{(1-1/C)a/2}{d/2+a/(2C)}\right)^{d/2+a/(2C)}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}<e^{(1-1/C)a/2}\cdot \left(\frac{d+a}{d+a/C}\right)^{(1-1/C) a/2}$

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 $\frac{d+a/C}{d+a}<1$, it's sufficient to prove

$\left(\frac{e}{3}C^{2\epsilon/a}\right)^C \cdot e^{-1} \cdot d^{(1+\epsilon)2C/a}\stackrel{?}{\le}1$

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 $e^{(1+\epsilon)2/K}\cdot \frac{e}{3}C^{2\epsilon}<1$.

REMARK. If one constrains $a$ slightly differently (that is: $\space a>K\log(d)+K$) then $\frac{e}{3}C^{2\epsilon/a}<1$ for $K$ large enough (without need to reduce $\epsilon$). This proves that for any $\epsilon>0$ there are $C$ and $K$ for which the inequality holds.