# How to prove that for all non-negative $\forall x\in\mathbb R: x\ge \frac{\ln^2(1+x+\sqrt{2x})}{2}$?

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?

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

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 $$x\geq0$$.

• Perfect, elegant and simple, thanks! – John D Jun 10 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!