Prove that $\prod\limits_{\mathrm{cyc}}\left(1+\frac{1}{\sqrt{ab}}\right)^a\geq 2^{a+b+c+d}$ for $a+b+c+d \le 4$ 
Let $a,b,c,d$ be positive real numbers satisfying $a+b+c+d\leq 4$. Show
$$\left(1+\frac{1}{\sqrt{ab}}\right)^a\left(1+\frac{1}{\sqrt{bc}}\right)^b
\left(1+\frac{1}{\sqrt{cd}}\right)^c\left(1+\frac{1}{\sqrt{da}}\right)^d\geq
2^{a+b+c+d}.$$

My attempt: I tried using the generalized GM-HM inequality,
$$RHS\geq \left(\frac{a+b+c+d}{\sum_{cyc}\frac{a\sqrt{ab}}{1+\sqrt{ab}}}\right)^{a+b+c+d}$$
and it remains to show that
$$a+b+c+d\geq 2\sum_{cyc}\frac{a\sqrt{ab}}{1+\sqrt{ab}}.$$
Any advice from here? Any idea for a different approach is also very welcome!
 A: Remarks: Here is a proof. The proof of (5) is not nice. Hope to see a nice proof.

Taking logarithm, the desired inequality is written as
$$\sum_{\mathrm{cyc}} a\ln \frac{1 + \frac{1}{\sqrt{ab}}}{2}
\ge 0. \tag{1}$$
Using $\ln x \ge \frac{2(x-1)}{1+x}$ for all $x > 0$ (easy to prove), we have
$$\ln \frac{1 + \frac{1}{\sqrt{ab}}}{2}
\ge \frac{8}{3} \cdot \frac{1}{3\sqrt{ab} + 1} - \frac23. \tag{2}$$
From (1) and (2), it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab} + 1} - \frac23(a + b + c + d) \ge 0$$
or
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}(a+b+c+d) + (a+b+c+d)} \ge \frac23. \tag{3}$$
Since $a+b+c+d \le 4$, from (3), it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}\cdot 4 + (a+b+c+d)} \ge \frac23. \tag{4}$$
Since the inequality (4) is homogeneous, WLOG, assume that $a + b + c + d = 4$. It suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}\cdot 4 + 4} \ge \frac23$$
or
$$\sum_{\mathrm{cyc}} \frac{a}{3\sqrt{ab} + 1} \ge 1. \tag{5}$$
A proof of (5) is given at the end.
We are done.

Proof of (5):
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sum_{\mathrm{cyc}} \frac{a}{3\sqrt{ab} + 1}
\ge \frac{\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2}{\sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 (3\sqrt{ab} + 1)}. \tag{6}$$
From (5) and (6), it suffices to prove that
$$\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2 \ge \sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 (3\sqrt{ab} + 1).$$
After homogenization, it suffices to prove that
$$\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2 \ge \sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 \left(3\sqrt{ab} + \frac{a + b + c + d}{4}\right).$$
Letting $a = x^2, b = y^2, c = z^2, d = w^2$, it suffices to prove that, for all $x, y, z, w \ge 0$,
$$\left[\sum_{\mathrm{cyc}} x (x + w)\right]^2 \ge \sum_{\mathrm{cyc}} (x + w)^2 \left(3xy + \frac{x^2 + y^2 + z^2 + w^2}{4}\right). \tag{7}$$
WLOG, assume that $w = \min(x, y, z, w)$.
Let $s = z - w, t = y - w, r = x - w$. Then $s, t, r \ge 0$. The inequality (7) is written as
$$A w^2 + Bw + C \ge 0 \tag{8}$$
where
\begin{align*}
 A &= 4\, \left( r-t \right) ^{2}+4\, \left( s-t \right) ^{2}+4\,{r}^{2}+4\,
 {s}^{2}, \\[6pt]
 B &= {\frac { \left( 3\,{s}^{2}-2\,st+3\,{t}^{2} \right) {r}^{3}}{{s}^{2}-2
   \,st+3\,{t}^{2}}}+{\frac {r \left( rs-3\,rt+2\,{s}^{2}-4\,st+6\,{t}^{2
   } \right) ^{2}}{{s}^{2}-2\,st+3\,{t}^{2}}}\\[8pt]
&\qquad +4\,{s}^{3}+3\,{s}^{2}t+t
 \left( 3\,s-2\,t \right) ^{2},\\[6pt]
 C &= \frac14\,{r}^{2} \left( 2\,r-3\,t \right) ^{2}+\frac18\, \left( 4\,s-3\,t
 \right) ^{2}{r}^{2}+\frac58\,{t}^{2}{r}^{2}+ \left( 3\,{s}^{2}t-8\,s{t}^{
  2}+3\,{t}^{3} \right) r\\
 &\qquad +{s}^{4}+3\,{s}^{3}t+4\,{s}^{2}{t}^{2}-3\,s{t}^
 {3}+{t}^{4}.
\end{align*}
It is easy to prove that $A, B, C \ge 0$.
Thus, the inequality (8) is true.
We are done.
