The required inequality
$$\frac{xyzt}{(x+y)(z+t)} \le \frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}$$
is equivalent to the inequality
$$\frac{(x+y)(z+t)}{xyzt} \ge \frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2}$$
(1)
Note that
$$\frac{2(x+y+z+t)}{(x+z)(y+t)} = \frac{2}{x+z} + \frac{2}{y+t}$$
By A.M. - G.M. inequality,
$$\frac{2}{x+z} \le \frac{1}{\sqrt{xz}}, \quad \frac{2}{y+t} \le \frac{1}{\sqrt{yt}}$$
hence
$$\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2} \le \left( \frac{1}{\sqrt{xz}} + \frac{1}{\sqrt{yt}} \right)^2$$
Further,
$$\frac{(x+y)(z+t)}{xyzt} = \frac{xz+xt+yz+yt}{xyzt}$$
$$= \frac{xz}{xyzt} + \frac{xt}{xyzt} + \frac{yz}{xyzt} + \frac{yt}{xyzt}$$
$$= \frac{1}{yt} + \frac{1}{yz} + \frac{1}{xt} + \frac{1}{xz}$$
But we can also write
$$\frac{(x+y)(z+t)}{xyzt} = \frac{xz+xt+yz+yt}{xyzt} = \frac{xz+yt+xt+yz}{xyzt}$$
$$= \frac{xz+yt}{xyzt} + \frac{xt+yz}{xyzt}$$
$$= \frac{xz}{xyzt} + \frac{yt}{xyzt} + \frac{xt}{xyzt} + \frac{yz}{xyzt}$$
$$= \frac{1}{yt} + \frac{1}{xz} + \frac{1}{xt} + \frac{1}{yz}$$
But the key step is to use the AM-GM inequality:
$$\frac{(x+y)(z+t)}{xyzt} \ge \left( \frac{1}{\sqrt{xz}} + \frac{1}{\sqrt{yt}} \right)^2$$
From the previous step,
$$\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2} \le \left( \frac{1}{\sqrt{xz}} + \frac{1}{\sqrt{yt}} \right)^2$$
Therefore,
$$\frac{(x+y)(z+t)}{xyzt} \ge \left( \frac{1}{\sqrt{xz}} + \frac{1}{\sqrt{yt}} \right)^2 \ge \frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2}$$
which proves the required inequality.