Prove $\frac{2}{\alpha+\beta} < \frac{\log \beta - \log \alpha}{\beta-\alpha} < \frac{1}{\sqrt{\alpha \beta}}$ for $0<\alpha<1<\beta$

Question

I found that, to prove some result in my thesis, I need the following inequality:

$\frac{2}{\alpha+\beta} < \frac{\log \beta - \log \alpha}{\beta-\alpha} < \frac{1}{\sqrt{\alpha \beta}}$ for $0<\alpha<1<\beta$.

However, I am having difficulties in proving the above inequality. Could someone prove this inequality? Thank you.

Asked and answered e.g. here: https://math.stackexchange.com/q/1084528/42969 — Martin R, Apr 19 '21 at 07:35
Does this answer your question? How to prove that $\log(x)<x$ when $x>1$? — Learnmore, Apr 23 '21 at 07:49

score 2 · Accepted Answer · 2021-04-19T07:27:21.010

2

Equivalently: $\dfrac{\log t}{2} > \dfrac{t-1}{t+1}, t > 1$. You can prove this one by consider $f(t) = (t+1)\log t - 2(t-1), t > 1$. It has $f'(t) = \log t + \dfrac{1}{t} -1 > 0$ since $f''(t) = \dfrac{t-1}{t^2} > 0\implies f(t) > f(1) = 0$, and this means the first inequality is proved. For the second, do the same sub, let $t = \dfrac{\beta}{\alpha} > 1\implies \log t < \sqrt{t} - \dfrac{1}{\sqrt{t}}$. This is done similarly and is left as a real exercise for you.

edited Apr 19 '21 at 07:27

answered Apr 19 '21 at 07:18

1

I really appreciate your help! – ssb Apr 19 '21 at 07:27

Prove $\frac{2}{\alpha+\beta} < \frac{\log \beta - \log \alpha}{\beta-\alpha} < \frac{1}{\sqrt{\alpha \beta}}$ for $0<\alpha<1<\beta$

1 Answers1