Monotonicity of function proof: Prove that $e^{\arctan x}(x+\sqrt{1+x^2}) < e^{2x}$ for all $x >0$

Question

I need to prove that

$$e^{\arctan x}(x+\sqrt{1+x^2}) < e^{2x}$$ for all $x >0$.

I proceeded as below: Let $$f(x) = e^{\arctan x}(x+\sqrt{1+x^2}) - e^{2x},$$ then I differentiated $f(x)$ and tried to establish that $f'(x) < 0$ for all $x >0$: $$f'(x)=e^{\arctan x}\frac{(1+x)(1+\sqrt{1+x^2})+x^2}{1+x^2}- 2e^{2x}.$$

Now, how to establish that this expression is negative for all positive $x$? It seems to be a very complex espression.

Answer 1

Hint: Prove that the function $$f(x)=e^{\arctan x-2x}(x+\sqrt{1+x^2})-1$$ is decreasing.

Edit for @AlexFrancisco: I write it step by step: \begin{align} f(x) &=e^{\arctan x-2x}(x+\sqrt{1+x^2})-1\\ f'(x) &=e^{\arctan x-2x}\left(\dfrac{-1-2x^2}{1+x^2}(x+\sqrt{1+x^2}) + \dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right)\\ &=e^{\arctan x-2x}(x+\sqrt{1+x^2})\left(\dfrac{-1-2x^2}{1+x^2} + \dfrac{\sqrt{1+x^2}}{1+x^2}\right)\\ &=e^{\arctan x-2x}(x+\sqrt{1+x^2})\dfrac{1}{1+x^2}\left( -1-2x^2+ \sqrt{1+x^2}\right) \end{align} where $e^{\arctan x-2x}>0$, $x+\sqrt{1+x^2}>0$, $\dfrac{1}{1+x^2}>0$ and $-1-2x^2+ \sqrt{1+x^2}<0$ because $1+x^2<(1+2x^2)^2$ or $x^2(4x^2+3)>0$.

Answer 2

You do not need any monotonicity argument: what you just need is the inequalities $\sqrt{1+t}< 1+\frac{t}{2}$ and $\arctan t< t$ for $t>0$. Both inequalities can be proved using functions given by formulas on left-hand-side are concave over $(0,\infty)$ and considering their tangent at $t=0$.

Then the remaining proof is easy, as $x+\sqrt{1+x^2} < 1 + x + \frac{x^2}{2} < e^x$.

Answer 3

To prove that $\;\mathrm e^{\arctan x}(x+\sqrt{1+x^2}) < \mathrm e^{2x}$, you can prove the same inequality for the logs, i.e. $$\arctan x + \ln\Bigl(x+\sqrt{1+x^2}\Bigr)=\arctan x +\operatorname{argsinh} x < 2x. $$

By the following corollary of the Mean Value theorem; it is enough to compare the derivatives of both sides and their value at $x=0$:

Let $f, g$ be differentiable functions defined on an interval $I$, $x_0\in I$. If

• $f(x_0)\le g(x_0)$, and
• $f'(x) <g'(x)$ for all $x>x_0$,

then $f(x) <g(x$ for all $x>x_0$.

Let's check the hypotheses of this corollary:

• $\arctan 0+\operatorname{argsinh}0 =2\cdot 0$,
• $\displaystyle\frac 1{1+x^2}+\frac 1{\sqrt{1+x^2}}<1+1=2$.