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 $x\geq7$

Question

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.

Answer 1

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.