How to prove this ln inequality? I have the following inequality, which (supposedly) holds for every $x\in\mathbb{R}$:
$$
1+x\ln\left(x+\sqrt{1+x^{2}}\right)\geq\sqrt{1+x^{2}}
$$
I've been struggling to find some known inequalities involving logarithms that could be applied here (and lack of condition for $x$ to be non-negative doesn't help either). I would be very helpful for hints on how this should be approached.
 A: Just to illustrate the trick for differentiating in my comment, using the same $f(x)$ as Norbert but writing $y=\sqrt{1+x^2}$:$$f(x)=1+x\log(x+y)-y$$ and knowing that $$\frac{dy}{dx}=\frac x y$$ we compute:
$$f'(x) = 0 + \log(x+y) +\frac {x (1+\frac x y)} {x+y} -\frac x y $$
And simplify ... which works neatly essentially for the same reason that Davide Giraudo's substitution works, to give first and second derivatives which are easy to work with:

So we get $$f'(x) = \log(x+y)$$ and $$f''(x)=\frac {1+\frac x y} {x+y}=\frac 1 y$$

We find that there is a minimum at $x=0$ with  $f(0)=0\text{ and } f''(x) \text { positive throughout, which gives us what we need ...}$
A: Hint 1: Set $f(x)=1+x \log(x+\sqrt{1+x^2})-\sqrt{1+x^2}$
Hint 2: Find stationary points from equation $f'(x)=0$.
Hint 3: Check that $f''(x)>0$ for all $x\in\mathbb{R}$
Hint 4: Conclude that $f$ attains its minimum, check that it equals $0$.
A: Write $x=\sinh t$. Then we have to proove that $1+\sinh t\ln(\sinh t+\cosh t)\geq \cosh t$, that is 
$$\forall t\in\Bbb R,1+\sinh t\times t\geq \cosh t.$$
Since 
$$1-\cosh t=-\frac 12\left(e^{t/2}-e^{—t/2}\right)^2=-2\sinh^2\frac  t2,$$
we have to show that 
$$\forall t\in\Bbb R,\sinh t\times t\geq 2\sinh^2\frac t2,$$
that is 
$$\forall t\geq 0, t\cdot \cosh\frac t2\geq\sinh\frac t2\Leftrightarrow\forall t\geq 0, f(t):=2t\cosh t-\sinh t\geq 0. $$
We have $f'(t)=\cosh t+2t\sinh t\geq 0$, hence $f$ is increasing. Since $f(0)=0$, we can conclude.
