I want to show that $$x+1 \neq (x^3(x+2))^{1/4} + \sqrt{x+1-\sqrt{x^2+2x}}$$ for any real $x>0$.

There are two approaches I've taken: showing they are equal and arriving at a contradiction (but this hasn't worked) and computing the derivative of the RHS and showing it is strictly less than $1$ everywhere. Its derivative is given by


I'm not sure how to show it is less than $1$. One approach may be to show that $g(x)$ is (strictly) increasing and $\lim_\limits{x\to \infty}g(x)=1$. Nothing has worked yet. Thank you for any help!

  • $\begingroup$ I assume you look for the real domain only. Use first the two square roots to limit x. Than do it with the 1/4 power. $\endgroup$ – Moti Jun 11 at 20:20
  • $\begingroup$ you can not have negative values under the square roots... $\endgroup$ – Moti Jun 12 at 2:43
  • 1
    $\begingroup$ (In less words, $(x+1)^2 \ge 1$ for any $x> 0$, and $\sqrt{x^2+2x} = \sqrt{(x+1)^2 - 1} \le \sqrt{(x+1)^2 } = x+1 $.) $\endgroup$ – Calvin Khor Jun 13 at 17:28

Suppose that there exists an $x\gt 0$ such that $$x+1 - \sqrt{x+1-\sqrt{x^2+2x}}= (x^3(x+2))^{1/4} $$

Squaring the both sides gives $$(x+1)^2-2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}+x+1-\sqrt{x^2+2x}=x\sqrt{x^2+2x},$$ i.e. $$(x+1)^2+x+1-(x+1)\sqrt{x^2+2x}=2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}$$

Let $x+1=y\ (\gt 1)$. Then, we have $$y^2+y-y\sqrt{y^2-1}=2y\sqrt{y-\sqrt{y^2-1}}$$ Dividing the both sides by $y$ gives $$y+1-\sqrt{y^2-1}=2\sqrt{y-\sqrt{y^2-1}}$$ Squaring the both sides gives $$(y+1)^2-2(y+1)\sqrt{y^2-1}+y^2-1=4\left(y-\sqrt{y^2-1}\right),$$ i.e. $$y^2+2y+1+y^2-1-2(y+1)\sqrt{y^2-1}=4y-4\sqrt{y^2-1},$$ i.e. $$2y^2+2y-4y=(2y+2-4)\sqrt{y^2-1},$$ i.e. $$2y(y-1)=2(y-1)\sqrt{y^2-1}$$ Dividing the both sides by $2(y-1)$ gives $$y=\sqrt{y^2-1}$$ which is impossible.

  • $\begingroup$ This is a really awesome proof! Thank you. I actually do not follow your final move after the "i.e." however. I got instead the following: $y^2+2y+1- 2y\sqrt{y^2-1} - 2\sqrt{y^2-1}+y^2-1=4y-4\sqrt{y^2-1}$ and hence $\sqrt{y^2-1}(2y+2)=2\sqrt{y^2-1}$ so that $y+1=1$ or $y=0$ contradicting that $y>1$. $\endgroup$ – Squirtle Jun 13 at 20:30
  • $\begingroup$ @Squirtle: I've added some steps. $\endgroup$ – mathlove Jun 14 at 4:02

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