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?

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

2 Answers 2


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


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


which simplifies to


which is certainly true for $x\geq0$.

  • 1
    $\begingroup$ Perfect, elegant and simple, thanks! $\endgroup$
    – John D
    Jun 10, 2019 at 22:35

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!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .