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I'm trying to prove that for all non-negative $\forall x\in\mathbb R:$ $$x\ge \frac{\ln^2(1+x+\sqrt{2x})}{2}.$$

You can think of it as a tighter inequality than the useful $x\ge \ln(1+x)$ or $e^x\ge 1+x$.

Using the Taylor expansion of $\ln(1+y)$, for $y=x+\sqrt{2x}$, gets ugly fast and it's hard to make anything out of it.

Any other ideas?

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    $\begingroup$ I meant $(\ln())^2$. $\endgroup$
    – John D
    Jun 10, 2019 at 22:28

2 Answers 2

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This can be written as

$$\sqrt{2x}\geq\ln(1+x+\sqrt{2x}), \quad x\geq0$$

raising both sides to the exponential function, the relation becomes

$$\text{e}^{\sqrt{2x}}\geq1+x+\sqrt{2x}$$

using the taylor expansion for the exponential function (and specifically writing out terms which will cancel with those on the right side), we have

$$1+\sqrt{2x}+\frac{2x}{2}+\sum_{n=3}^\infty\frac{(2x)^{n/2}}{n!}\geq1+x+\sqrt{2x}$$

which simplifies to

$$\sum_{n=3}^\infty\frac{(2x)^{n/2}}{n!}\geq0$$

which is certainly true for $x\geq0$.

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    $\begingroup$ Perfect, elegant and simple, thanks! $\endgroup$
    – John D
    Jun 10, 2019 at 22:35
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Let $\sqrt{2x}=t$.

Thus, $t\leq0$, $\ln\left(1+t+\frac{t^2}{2}\right)\geq0$ and we need to prove that $$t^2\geq\ln^2\left(1+t+\frac{t^2}{2}\right)$$ or $f(t)\geq0,$ where $$f(t)=t-\ln\left(1+t+\frac{t^2}{2}\right).$$ But, $$f'(t)=1-\frac{1+t}{1+t+\frac{t^2}{2}}\geq0,$$ which says $$f(t)\geq f(0)=0$$ and we are done!

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