On some inequality (upper bound) on a function of two variables There is a problem (of physical origin) which needs an analytical solution or a hint.  Let us consider the following real-valued function of two variables
$y (t,a) = 4 \left(1 + \frac{t}{x(t,a)}\right)^{ - a - 1/2}  
  \left(1 - \frac{2}{x(t,a)}\right)^{ 1/2} (x(t,a))^{-3/2} (z(t,a))^{1/4} \left(1 + \frac{t}{2}\right)^{ a},$
where  $t  > 0$,  $0 < a < 1$ and
$x(t,a)   =   \frac{a-1}{2} t + \frac{3}{2}+\frac{1}{2}\sqrt{z(t,a)}$,
$z(t,a)   =   (1-a)^2 t^2+ 2(3-a)t +9.$
It is necessary to prove  that
$$y(t,a) < 1$$
for all $t  > 0$ and $0 < a < 1$. The numerical analysis supports this bound.
P.S. I apologize for too ``technical'' question. It looks that the inequality is valid also for $a = 1$
but it fails for $a > 1$.

a human verifiable proof:
Let us prove that the sharp bound of $y$ is $\frac{\sqrt{3+2\sqrt 3}}{3}$.
Letting $x := x(t, a), z := z(t, a), y := y(t, a)$, we have
$$y = 4 \left(\frac{1 + t/2}{1 + t/x}\right)^a\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/4} (1 - 2/x)^{1/2} x^{-3/2}. \tag{1}$$
Using Bernoulli inequality $(1 + u)^r \le 1 + ru$ for all $r \in (0, 1]$ and all $u > -1$, we have
$$y \le 4 \left(1 + \left(\frac{1 + t/2}{1 + t/x}  - 1\right)a\right)\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/4} (1 - 2/x)^{1/2} x^{-3/2} \tag{2}$$
and
$$y^2 \le 16 \left(1 + \left(\frac{1 + t/2}{1 + t/x}  - 1\right)a\right)^2\cdot \left(\frac{z}{(1 + t/x)^2}\right)^{1/2} (1 - 2/x) x^{-3}. \tag{3}$$
We first simplify the expression. We have
$$a \in (0, 1) \quad \iff \quad a = \frac{1}{1 + s}, \quad s > 0. $$
Then, we have (the so-called Euler substitution)
$$t > 0 \quad \iff \quad t = \frac{2(1+s)(3s+2 - 3u))}{u^2 - s^2}, \quad s < u < s + \frac23. $$
With these substitutions, we have
$$z = \left(\frac{-3u^2 + (6s + 4)u - 3s^2}{u^2 - s^2}\right)^2$$
and
$$x = \frac{6s + 2}{u + s}.$$
Then, (3) is written as
$$y^2 \le {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right)
        \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}}.
$$
The constraints are: $s > 0$ and $s < u < s + \frac23$.
We have
$$\sup_{s> 0, ~ s < u < s + \frac23} {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right)
        \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}} = \frac{3 + 2\sqrt 3}{9}. \tag{4}$$
(The proof is given at the end.)
Thus, we have
$$y \le \frac{\sqrt{3 + 2\sqrt 3}}{3}.$$
On the other hand, in (1), letting
$t = 3 + 3\sqrt 3, ~ a = 1$,
we have $y = \frac{\sqrt{3 + 2\sqrt 3}}{3}$.
Thus, $\frac{\sqrt{3 + 2\sqrt 3}}{3}$ is a sharp bound.
We are done.

Proof of (4):
Fixed $s > 0$, let
$$f(u) := {\frac { \left(-3u^2 + 6su + 4u - 3s^2 \right)
        \left( 5\,s+2 - u \right) ^{2}}{4 \left( 3\,s+1 \right) ^{3}}}.$$
The maximum of $f(u)$ is attained
at $u_0 = 2\,s+1- \frac13\,\sqrt {9\,{s}^{2}+12\,s+3}$. Let
$g(s) := f(u_0)$.
It is easy to prove that $g'(s) < 0$ for all $s \ge 0$.
Thus, $g(s) \le g(0) =  \frac{3 + 2\sqrt 3}{9}$.
We are done.
