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.

  • $\begingroup$ Try taking everything over to the left hand side and differentiate the result. $\endgroup$ – Mark Bennet Aug 13 '12 at 20:38
  • $\begingroup$ You can simplify the manipulation rather by setting $y=\sqrt{1+x^2}$ and noting that $\frac {dy}{dx} = \frac x y$ $\endgroup$ – Mark Bennet Aug 13 '12 at 20:45

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 $f(0)=0\text{ and } f''(x) \text { positive throughout, which gives us what we need ...}$

  • $\begingroup$ OK, now I understand your method too. Thank you! $\endgroup$ – Johnny Westerling Aug 13 '12 at 21:45

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 $0$.

  • $\begingroup$ Thank you for the answer. Although your method gives us a clear and certain answer, because of the sheer length of the second-order derivative of $f$, I assume that there is some quicker (though maybe less intuitive) way to prove this. $\endgroup$ – Johnny Westerling Aug 13 '12 at 21:00

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.

  • $\begingroup$ Thanks! I understand it, until the moment where we move from $\forall_{t\in\mathbb{R}},\sinh{t}\cdot{t}\geq{2}\sinh^{2}\frac{t}{2}$ to the next line. Could you please explain this transition? $\endgroup$ – Johnny Westerling Aug 13 '12 at 21:32
  • $\begingroup$ I replaces $1-\cosh t$ by what I computed in the last line. $\endgroup$ – Davide Giraudo Aug 13 '12 at 21:33
  • $\begingroup$ I think I understand your method as well... Thanks! $\endgroup$ – Johnny Westerling Aug 13 '12 at 22:04

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