WLOG, assume that $d = \min(a,b,c,d)$. Note that $x\mapsto \sqrt{x}$ for $x\ge 0$ is concave. We have
\begin{align}
&\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}}\\
=\ & \frac{a}{a+b}\sqrt{a+b} + \frac{8b}{11(b+c)}\sqrt{\frac{121}{64}(b+c)}\\
\le\ & \Big(\frac{a}{a+b} + \frac{8b}{11(b+c)}\Big)
\sqrt{\frac{\frac{a}{a+b}}{\frac{a}{a+b} + \frac{8b}{11(b+c)}}(a+b)
+ \frac{\frac{8b}{11(b+c)}}{\frac{a}{a+b} + \frac{8b}{11(b+c)}}\frac{121}{64}(b+c) }\\
=\ & \sqrt{\frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}}
\end{align}
and
\begin{align}
&\frac{c}{\sqrt{c+d}} + \frac{d}{\sqrt{d+a}}\\
=\ & \frac{c}{c+d}\sqrt{c+d} + \frac{d}{d+a}\sqrt{d+a}\\
\le \ & (\frac{c}{c+d} + \frac{d}{d+a})\sqrt{\frac{\frac{c}{c+d}}{\frac{c}{c+d} + \frac{d}{d+a}}(c+d)
+ \frac{\frac{d}{d+a}}{\frac{c}{c+d} + \frac{d}{d+a}}(d+a) }\\
=\ & \sqrt{\frac{ca+2cd+d^2}{d+a}}.
\end{align}
Thus, it suffices to prove that
$$\sqrt{\frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}} + \sqrt{\frac{ca+2cd+d^2}{d+a}} \le \frac{\sqrt{33}}{4}\sqrt{a+b+c+d}.$$
Denote $A = \frac{(8a+11b)(19ab+11ca + 8b^2)}{88(a+b)(b+c)}$ and $B = \frac{ca+2cd+d^2}{d+a}$. Squaring both sides of the inequality, it suffices to prove that
$$2\sqrt{AB} \le \frac{33}{16}(a+b+c+d) - A - B.$$
Note that $\frac{33}{16}(a+b+c+d) - A - B = \frac{1}{176(a+b)(b+c)(d+a)} f(a, b, c, d)$
where $f(a,b,c,d)$ is a polynomial with non-negative coefficients.
Thus, it suffices to prove that
$$\Big(\frac{33}{16}(a+b+c+d) - A - B\Big)^2 - 4AB \ge 0$$
or
$$1024g(a,b,c,d) + h(a,b,c,d)\ge 0$$
where $h(a,b,c,d)$ is a polynomial with non-negative coefficients, and $g(a,b,c,d)$ is a polynomial given by
$$
g(a,b,c,d) = 3\, a^6\, b^2 + 21\, a^6\, b\, c + 34\, a^6\, c^2 + 20\, a^5\, b^3 - 66\, a^5\, b^2\, c + 48\, a^5\, b^2\, d - 64\, a^5\, b\, c^2 + 217\, a^5\, b\, c\, d - 53\, a^5\, c^3 + 200\, a^5\, c^2\, d + 53\, a^4\, b^4 - 217\, a^4\, b^3\, c + 210\, a^4\, b^3\, d - 144\, a^4\, b^2\, c^2 + 416\, a^4\, b^2\, c\, d - 14\, a^4\, b^2\, d^2 + 25\, a^4\, b\, c^3 + 199\, a^4\, b\, c^2\, d + 213\, a^4\, b\, c\, d^2 + 34\, a^4\, c^4 - 159\, a^4\, c^3\, d + 242\, a^4\, c^2\, d^2 + 65\, a^3\, b^5 - 58\, a^3\, b^4\, c + 366\, a^3\, b^4\, d + 192\, a^3\, b^3\, c^2 + 25\, a^3\, b^3\, c\, d - 50\, a^3\, b^3\, d^2 + 279\, a^3\, b^2\, c^3 - 406\, a^3\, b^2\, c^2\, d + 401\, a^3\, b^2\, c\, d^2 - 55\, a^3\, b^2\, d^3 + 68\, a^3\, b\, c^4 - 330\, a^3\, b\, c^3\, d + 365\, a^3\, b\, c^2\, d^2 + 25\, a^3\, b\, c\, d^3 + 4\, a^3\, c^4\, d - 162\, a^3\, c^3\, d^2 + 79\, a^3\, c^2\, d^3 + 34\, a^2\, b^6 + 79\, a^2\, b^5\, c + 264\, a^2\, b^5\, d + 242\, a^2\, b^4\, c^2 - 216\, a^2\, b^4\, c\, d + 79\, a^2\, b^4\, d^2 + 200\, a^2\, b^3\, c^3 - 302\, a^2\, b^3\, c^2\, d + 167\, a^2\, b^3\, c\, d^2 - 232\, a^2\, b^3\, d^3 + 34\, a^2\, b^2\, c^4 - 23\, a^2\, b^2\, c^3\, d - 186\, a^2\, b^2\, c^2\, d^2 - 6\, a^2\, b^2\, c\, d^3 + 34\, a^2\, b^2\, d^4 + 8\, a^2\, b\, c^4\, d - 408\, a^2\, b\, c^3\, d^2 + 230\, a^2\, b\, c^2\, d^3 + 68\, a^2\, b\, c\, d^4 + 4\, a^2\, c^3\, d^3 + 34\, a^2\, c^2\, d^4 + 68\, a\, b^6\, d - 26\, a\, b^5\, c\, d + 145\, a\, b^5\, d^2 + 111\, a\, b^4\, c^2\, d - 135\, a\, b^4\, c\, d^2 - 230\, a\, b^4\, d^3 + 148\, a\, b^3\, c^3\, d - 414\, a\, b^3\, c^2\, d^2 + 53\, a\, b^3\, c\, d^3 + 6\, a\, b^3\, d^4 + 4\, a\, b^2\, c^4\, d - 238\, a\, b^2\, c^3\, d^2 + 291\, a\, b^2\, c^2\, d^3 + 136\, a\, b^2\, c\, d^4 + 8\, a\, b\, c^3\, d^3 + 68\, a\, b\, c^2\, d^4 + 34\, b^6\, d^2 - 106\, b^5\, c\, d^2 - 53\, b^5\, d^3 - 102\, b^4\, c^2\, d^2 + 83\, b^4\, c\, d^3 + 34\, b^4\, d^4 + 8\, b^3\, c^3\, d^2 + 140\, b^3\, c^2\, d^3 + 68\, b^3\, c\, d^4 + 4\, b^2\, c^3\, d^3 + 34\, b^2\, c^2\, d^4.$$
It suffices to prove that $g(a,b,c,d)\ge 0$ for $d = \min(a,b,c,d)$. I verify it numerically. Maybe someone can give nice proofs of it.
