calculus - How to prove that for all non-negative $forall xinmathbb R: xge frac{ln^2(1+x+sqrt{2x})}{2}$?

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?

Answer

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 $x\geq0$.