19 Find all positive real numbers $a$, such that for any positive real numbers $t_{1}, t_{2}, t_{3}, t_{4}$ satisfying $t_{1} \cdot t_{2} \cdot t_{3} \cdot t_{4}=a^{4}$, we have
$$
\frac{1}{\sqrt{1+t_{1}}}+\frac{1}{\sqrt{1+t_{2}}}+\frac{1}{\sqrt{1+t_{3}}}+\frac{1}{\sqrt{1+t_{4}}} \leqslant \frac{4}{\sqrt{1+a}} \text {. }
$$
Let $t_{1}=t_{2}=t_{3}=t, t_{4}=\frac{a^{4}}{t^{3}}$.
Let $t \rightarrow 0^{+}$, we get $3 \leqslant \frac{4}{\sqrt{1+a}}$, hence $a \leqslant \frac{7}{9}$.
Conversely, when $00)$ the maximum value is
$$
f_{\max }=\left\{\begin{array}{ll}
\frac{2}{\sqrt{1+\sqrt{A}}}, & \text { when } 04 \text { . }
\end{array}\right.
$$

Proof: $f^{2}(x)=\frac{1}{1+x}+\frac{1}{1+\frac{A}{x}}+\frac{2}{\sqrt{(1+x)\left(1+\frac{A}{x}\right)}}$
$$
\begin{array}{l}
=\frac{2+x+\frac{A}{x}}{1+A+x+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\
=1+(1-A) u^{2}+2 u=g(u),
\end{array}
$$

where $\quad u=\frac{1}{\sqrt{1+A+x+\frac{A}{x}}} \in\left(0, \frac{1}{1+\sqrt{A}}\right]$.
When $04$, the axis of symmetry $\frac{1}{A-1}<\frac{1}{1+\sqrt{A}}$, the downward-opening quadratic function $g(u)$ reaches its maximum value $\frac{A}{A-1}$ at $u=\frac{1}{A-1}$.
Without loss of generality, assume $t_{1} \leqslant t_{2} \leqslant t_{3} \leqslant t_{4}$.
Let $A=t_{1} \cdot t_{2}, B=t_{3} \cdot t_{4}$, then
$$
A \leqslant a^{2}<4, A \cdot B=a^{4}, \sqrt{A} \cdot \sqrt{B}=a^{2},
$$

hence $\frac{1}{\sqrt{1+t_{1}}}+\frac{1}{\sqrt{1+t_{2}}} \leqslant \frac{2}{\sqrt{1+\sqrt{A}}}$.
Case one: If $B \leqslant 4$, then
$$
\frac{1}{\sqrt{1+t_{3}}}+\frac{1}{\sqrt{1+t_{4}}} \leqslant \frac{2}{\sqrt{1+\sqrt{B}}} \text {. }
$$

Thus
$$
\begin{aligned}
& \frac{1}{\sqrt{1+t_{1}}}+\frac{1}{\sqrt{1+t_{2}}}+\frac{1}{\sqrt{1+t_{3}}}+\frac{1}{\sqrt{1+t_{4}}} \\
\leqslant & \frac{2}{\sqrt{1+\sqrt{A}}}+\frac{2}{\sqrt{1+\sqrt{B}}} \leqslant \frac{4}{\sqrt{1+a}} .
\end{aligned}
$$

Case two: If $B \geqslant 4$, then
$$
\frac{1}{\sqrt{1+t_{3}}}+\frac{1}{\sqrt{1+t_{4}}} \leqslant \sqrt{\frac{B}{B-1}} .
$$

Let $t=\frac{1}{\sqrt{B}}$, then $0<t \leqslant \frac{1}{2}, \sqrt{A}=\frac{a^{2}}{\sqrt{B}}=a^{2} t$,
$$
\begin{aligned}
& \frac{1}{\sqrt{1+t_{1}}}+\frac{1}{\sqrt{1+t_{2}}}+\frac{1}{\sqrt{1+t_{3}}}+\frac{1}{\sqrt{1+t_{4}}} \\
\leqslant & \frac{2}{\sqrt{1+a^{2} t}}+\frac{1}{\sqrt{1-t^{2}}}=h(t), \\
h^{\prime}(t)= & {\left[2\left(1+a^{2} t\right)^{-\frac{1}{2}}+\left(1-t^{2}\right)^{-\frac{1}{2}}\right]^{\prime} } \\
= & -a^{2}\left(1+a^{2} t\right)^{-\frac{3}{2}}+t\left(1-t^{2}\right)^{-\frac{3}{2}}
\end{aligned}
$$

is strictly increasing on $\left(0, \frac{1}{2}\right)$, and $h^{\prime}(0)=-a^{2}<0$, so $h(t)$ is strictly decreasing or first strictly decreasing then strictly increasing on $\left[0, \frac{1}{2}\right]$. Also, $h(0)=3 \leqslant \frac{4}{\sqrt{1+a}}, h\left(\frac{1}{2}\right) \leqslant$ $\frac{4}{\sqrt{1+a}}$. Therefore,
$h(t) \leqslant \max \left\{h(0), h\left(\frac{1}{2}\right)\right\} \leqslant \frac{4}{\sqrt{1+a}}$ (i.e., the case when $B=4$ in Case one).
Therefore, the required positive real numbers $a$ are $\left(0, \frac{7}{9}\right)$.