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I am trying to prove that $\log(1+x) \leq \sqrt{x}$ for all values $x>x_0$ which implies that $\log(1+x) = O(\sqrt{x})$. However, I am not able to get to this result. The result which states $\log(1+x) = O(x)$ is immediate as $\log(1+x) \leq x, ~\forall x\geq0$ which follows by expanding $\log(1+x)$ around $0$. However, I cannot show that $\log(1+x)\leq \sqrt{x}$ using the same approach.

Any help would be greatly appreciated!

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For $x>1$ we have $1/x < 1/\sqrt{x}$. Integrating both sides will get you a relationship between the logarithm and $\sqrt{x}$ that can be massaged to get the desired result.

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    $\begingroup$ Thanks for the comment. After your hint, I instantly derived it! $\endgroup$ Dec 27 '19 at 4:19
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$\ln (1+x)=\sqrt x$ when $x=0$. Furthermore, when $x\ge0$, $\dfrac d{dx} ln(1+x)\le \dfrac d{dx} \sqrt x $ because $\dfrac1{1+x} \le \dfrac1{2\sqrt x}$ because $2\sqrt x \le 1+x$ because $0\le(1-\sqrt x)^2$.

Therefore, $\ln(1+x)\le \sqrt x$ when $x\ge0$.

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  • $\begingroup$ Thank you for your comment :) $\endgroup$ Dec 27 '19 at 4:24
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The inequality you stated:

$$\ln(x)\le x-1\le x$$

Substitute $x\mapsto\sqrt[4]x$ and use log rules to get:

$$\ln(x)\le4\sqrt[4]x$$

But for $x\ge4^4$ we have

$$4\sqrt[4]x\le\sqrt x$$

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  • $\begingroup$ Thanks for your comment! This is also a very good way of showing it! $\endgroup$ Dec 27 '19 at 4:19

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