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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?

Funny
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  • I think the formula for the derivative is wrong. And why do you compare it to $0$? What you need to show is that $f(x)-x>0$. At $x=0$, $f(x)=0$. So taking the derivative you need to show that $f'(x)-1>0$, or $f'(x)>1$. – Andrei Oct 10 '20 at 18:23
  • Thank you so much! I will check the derivative – Funny Oct 10 '20 at 18:27

2 Answers2

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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

Raffaele
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  • Thank you very much! How have you calculated 2.51? – Funny Oct 10 '20 at 20:17
  • @Funny Solved the inequality with Wolfram Mathematica. Anyway you can see easily graphing $f(x)$ and $x^2$ that the inequality holds only on a bounded interval. – Raffaele Oct 10 '20 at 20:22
  • I appreciate your clear inequalities, however, I would also appreciate the way to prove your result without a graph – Funny Oct 10 '20 at 20:28
  • Can you solve $\frac{x}{\sqrt{x^2+1}}+\sqrt{x^2+1} \ln \left(x^2+1\right)>x^2$? – Raffaele Oct 10 '20 at 20:31
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We need to prove that: $$\sqrt{\frac{x}{x+1}}+\sqrt{1+x}\ln(1+x)>\sqrt{x},$$ where $x>0$, or $g(x)>0,$ where $$g(x)=\ln(1+x)-\frac{\sqrt{x}(\sqrt{x+1}-1)}{1+x}$$ and easy to show that $g'(x)>0,$ which gives $g(x)>g(0)=0$.

Michael Rozenberg
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