Showing $ 2\sqrt{\frac{x+3}{x}}+8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $ for $x\geq7$ 
Suppose that $x\geq 7$. I would like to show that
$$ 2\sqrt{\frac{x+3}{x}} + 8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $$

I rewrote the inequality as
$$ 2\sqrt{\frac{x+3}{x}} +8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{\left(\frac{x+1}{x}\right)^{3/2}\left(\frac{x+3}{x}\right)}{\left(\frac{x-1}{x}\right)^{5/2}}\right)\geq 10, $$
which is equivalent to
$$  2\sqrt{\frac{x+3}{x}} -2\ln\left( \sqrt{\frac{x+3}{x}}\right)+8\sqrt{\frac{x+1}{x}}-3\ln\left( \sqrt{\frac{x+1}{x}}\right) \geq 10-5\ln\left(\sqrt{\frac{x}{x-1}} \right). $$
I am not sure if rewriting it in the last way helps. Wolfram Alpha shows me that the inequality is true, but I cannot figure out why. Any help is greatly appreciated. Thank you.

We need to prove that, for all $x \ge 7$,
$$2\sqrt{\frac{x+3}{x}}-2\ln \sqrt{\frac{x+3}{x}} +8\sqrt{\frac{x+1}{x}}-3\ln \sqrt{\frac{x+1}{x}} \geq 10+5\ln\sqrt{\frac{x}{x-1}} .$$
With the substitution $x = \frac{1}{y}$,
it suffices to prove that, for all $y\in [0, 1/7]$,
$$2\sqrt{1 + 3y} - \ln(1 + 3y) + 8\sqrt{1 + y} - \frac32\ln(1 + y) \ge 10 - \frac52\ln(1 - y).$$
Let $F(y) = \mathrm{LHS} - \mathrm{RHS}$.
We have
$$F'(y) = \frac{3}{\sqrt{1 + 3y}}
- \frac{3}{1 + 3y} + \frac{4}{\sqrt{1 + y}}
- \frac{3}{2(1 + y)} - \frac{5}{2(1 - y)}$$
and
$$F''(y) = - \frac{9}{2(1 + 3y)^{3/2}} + \frac{9}{(1 + 3y)^2} - \frac{2}{(1 + y)^{3/2}} + \frac{3}{2(1 + y)^2} - \frac{5}{2(1 - y)^2}$$
and
$$F'''(y) = \frac{81}{4(1 + 3y)^{5/2}} - \frac{54}{(1 + 3y)^3} + \frac{3}{(1 + y)^{5/2}} - \frac{3}{(1 + y)^3}
- \frac{5}{(1 - y)^3}.$$
We have, for all $y\in [0, 1/7]$,
\begin{align*}
 F'''(y)
 &\le \left(\frac{81}{4(1 + 3y)^2} -  \frac{54}{(1 + 3y)^3}\right) + \left(\frac{3}{(1 + y)^2} - \frac{3}{(1 + y)^3}
 - \frac{5}{(1 + y)^3}\right)\\
 &= \frac{27(9y - 5)}{4(1 + 3y)^3} + \frac{3y - 5}{(1 + y)^3}\\
 & < 0.
\end{align*}
Note that $F''(0) > 0$ and $F''(1/7) < 0$.
Thus, there exists $y_0 \in (0, 1/7)$
such that $F''(y_0) = 0$,
$F''(y) > 0$ on $y\in [0, y_0)$,
and $F''(y) < 0$ on $y\in (y_0, 1/7]$.
Note that $F'(0) = 0$ and $F'(1/7) < 0$.
Thus, there exists $y_1 \in (y_0, 1/7)$ such that $F'(y_1) = 0$, $F'(y) > 0$ on $(0, y_1)$,
and $F'(y) < 0$ on $(y_1, 1/7)$.
Note that $F(0) = 0$ and $F(1/7) > 0$.
Thus, $F(y) \ge 0$ on $[0, 1/7]$.
We are done.
