Proof $f(x)>x$ for $x>0$ I'm a little stuck at proving that for these function:
$$\  f(x) = \frac{x}{\sqrt{1+x^2}} + \sqrt{1+x^2}\cdot\ln(1+x^2) $$
$f(x)>x$ for every $x>0$. Another question: is there any $a$ that $f(x)>x^2$ for every $x>a$?
My steps
I have computed the derivative:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) $$
then compared to zero:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) > 0 $$
and calculated that:
$$ 4>x^2(1+x^2)(x-\ln(1+x^2))^2 $$
Each of the components is $>0$ I can see from the primary function that the root is 0, but I cannot write it in more precise way than the equation above. I am thinking of Mean Value Theroem, but I cannot see, how to apply it in this case. I was considering something like:
$$ f'(c) = \frac{f(x) - f(0)}{x} = \frac{f(x)}{x} > 0 $$
Since the function gives positive values for $x>0$, but it has leaded me to nowhere. I would really appreciate your help.
EDIT
As Andrei suggested, I have corrected the derivative:
$$ \frac{x\ln(x^2+1)}{\sqrt{x^2+1}} + \frac{2x}{\sqrt{x^2+1}} + \frac{1}{(x^2+1)^{3/2}} $$
Then calculated:
$$ 1>(x^2+1)(\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
Since
$$ \sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 \\
x^2+1<4x^2+4x^2\ln(x^2+1)+x^2\ln(x^2+1)^2 $$
The positive and negative multiplication is smaller than 1, thus the inequality is true.
However, what about $f(x)>x^2$? Do I have to calculate the second derivative?
I have continued Andrei method and for $f'(x)-2x>0$ obtained:
$$ 1>(x^2+1)(2x\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
$$ 2x\sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 $$
And reduced to:
$$ x^4<\frac{3}{2}\ln(x^2+1) /^e(\cdot) \\
e^{x^4} < (x^2+1)\sqrt{x^2+1} $$
The order of growth indicates that this is false (only even power of x) and there is no such a that fulfil these requirements.
Is it correct?

For $x>0$
$$\frac{x}{\sqrt{x^2+1}}+\sqrt{x^2+1} \ln \left(x^2+1\right)>\frac{x}{\sqrt{x^2+1}}+\sqrt{x^2+1}>x;\forall x\in\mathbb{R}$$
Furthermore $f(x)>x^2$ only for $0<x<2.51$ (approximated), so there is no $a$ such that $f(x)>x^2$ for all $x>a$.
Hope this helps
