2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I meant $(\ln())^2$. $\endgroup$ – John D Jun 10 at 22:28
7
$\begingroup$

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 $x\geq0$.

$\endgroup$
  • 1
    $\begingroup$ Perfect, elegant and simple, thanks! $\endgroup$ – John D Jun 10 at 22:35
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.