Determine this function's minimum

Question

Let $f:\mathbb{R} \to \mathbb{R}$ with $$f(x)=\frac{x^2+1}{\sqrt{x^2+1}+x}$$ Determine its minimum.

I tried to use AM-GM in some form, but didn't get too much. I also thought about dividing both the numerator and denominator by some power of $x$ and then make a notation, or use derivatives, but both failed because of the square root...

Answer 1

Note that $$f’(x)= \frac{(\sqrt{x^2+1}+x)(2x)- (x^2+1)(\frac{x}{\sqrt{x^2+1}}+1)}{(\sqrt{x^2+1}+x)^2} = \frac{2x\sqrt{x^2+1}-x^2-1}{x\sqrt{x^2+1}+x^2+1}$$

Now, $f’(x)=0$ if $x^2=\frac{1}{3}$ which can be easily checked. Now, computing the second derivative, can you come to a conclusion?

Answer 2

Show that $$\frac{x^2+1}{\sqrt{x^2+1}+x}\geq \frac{4}{3\sqrt{3}}$$ the equal sign holds if $$x=\frac{1}{\sqrt{3}}$$

Answer 3

We may write $f(x)$ as $$f(x)= (x^2+1)\left(\sqrt{x^2+1}-x\right) $$ and notice that $$ f(\sinh\theta) = e^{\theta}\cosh^2\theta $$ implying $$ f(\sinh\log u) = \frac{u}{4}\left(u+\frac{1}{u}\right)^2. $$ The minimum of the RHS over $u>0$ is reached at $u=\frac{1}{\sqrt{3}}$ and the value of such a minimum is $\frac{4}{3\sqrt{3}}$. You can prove this statement by computing $\frac{d}{du}$, by the AM-GM inequality or by equating to zero the discriminant of the quartic polynomial $u^4+2u^2-4ku+1$. Since both $\sinh$ and $\log $ are continuous and increasing on their maximal domains, it follows that $$ \min_{x\in\mathbb{R}} f(x)=\frac{4}{3\sqrt{3}}.$$