How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$?

Question

How do I show this inequality that for $x>1$ $$\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$$

So far I have tried to use the inequality $\ln({x+1}) < x$, but it is not good enough.

For $x>1$, the inequality is equivalent to $\frac{\log(x+1)}{x} \le \frac{\log(x)}{x-1}$. Consider a decreasing function $f(x) = \frac{\log(x+1)}{x}$ . — dust05, Dec 22 '20 at 17:33

score 4 · Answer 1 · answered Dec 22 '20 at 17:36

4

You can show it using concavity of $\ln x$ as follows:

The given inequality is equivalent to

$$\ln x \geq \frac{x-1}{x}\ln(x+1)$$

Now, since $x>1$ we have

$$\frac{x-1}{x}\ln(x+1) + \frac 1x\ln 1 \stackrel{concavity}{\leq}\ln\left(\frac{x-1}{x}(x+1) +\frac 1x\cdot 1\right) =\ln x$$

Done.

answered Dec 22 '20 at 17:36

trancelocation

Nice solution, sometimes the answer is easy – test Dec 22 '20 at 17:38
1

@question : Thanks for your appreciation :-) – trancelocation Dec 22 '20 at 17:40

Quanto · Answer 2 · 2020-12-22T18:58:33.583

1

For $x>1$ $$x\ln x -(x-1)\ln(x+1) =\int_1^x \int_t^\infty \frac{u-1}{u(u+1)^2}\>dudt\ge0$$ Rearrange the inequality to obtain

$$\frac{\ln({x+1})}{\ln x} \leq \frac{x}{x-1}$$

edited Dec 22 '20 at 18:58

answered Dec 22 '20 at 17:56

Quanto

score 0 · Answer 3 · answered Dec 22 '20 at 17:31

(Fill in the gaps as needed. If you're stuck, show you work and explain what you've tried.)

Hint: Prove that $(x+1)^{x-1} < x^x $ for $ x > 1$.

Further hint: Direct application of weighted AM-GM.

Corollary: Inequality as stated, after taking logarithms.

3 Answers3