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Let $0<x<1$. Show that $$\ln{(\text{arcsinh}{(x)})}<\dfrac{1}{3}\ln{(\arcsin(x))}+\dfrac{2}{3}\ln{(\arctan(x))}$$

My try: since $$\Longleftrightarrow \text{arcsinh}{(x)}<\sqrt[3]{\arcsin(x)}\cdot\sqrt[3]{(\arctan(x))^2}$$

This inequality is very strong: see: http://www.wolframalpha.com/input/?i=ln%28arcsinhx%29-1%2F3ln%28arcsinx%29-2%2F3ln%28arctanx%29

then I can't, It is said this inequality can use Hölder's inequality:

$$\bigg|\int g(x)h(x)dx\bigg|\le\bigg[\int |g(x)|^pdx\bigg]^{\frac{1}{p}}\bigg[\int|h(x)|^qdx\bigg]^{\frac{1}{q}}$$ where $\dfrac{1}{p}+\dfrac{1}{q}=1$

http://en.wikipedia.org/wiki/Hölder's_inequality

because $$\dfrac{1}{3}+\dfrac{2}{3}=1$$

HK Lee
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math110
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1 Answers1

6

Let $$g(t) = \frac1{(1-t^2)^{1/6}}, \qquad h(t) = \frac1{(1+t^2)^{2/3}}$$

Then $$\int_0^xg^3 dt = \operatorname{asin}x, \qquad \int_0^x h^{3/2} dt = \operatorname{atan}x$$

So Hölder's inequality gives $$\int_0^xgh \, dt \le \left(\operatorname{asin} x\right)^{1/3}\cdot \left(\operatorname{atan}x\right)^{2/3} \tag{1}$$

Now note that for $0 < t < 1$,

$$gh = \frac1{\left(1-t^2\right)^{1/6} }\frac1{\left(1+t^2\right)^{2/3}} =\frac1{\left(1-t^4\right)^{1/6}}\frac1{\left(1+t^2\right)^{1/2}} > \frac1{\sqrt{1+t^2}}$$

hence we have $\displaystyle \int_0^x gh \, dt > \int_0^x \frac{dt}{\sqrt{1+t^2}} = \operatorname{asinh}x$

Using this in (1), we are done.

Macavity
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