# Showing that two curves do not intersect

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

$$g(x)=\frac{1-\frac{2x+2}{2\sqrt{x^2+2x}}}{2\sqrt{-\sqrt{x^2+2x}+x+1}}+\frac{x^3+3x^2\left(x+2\right)}{4\left(x^3\left(x+2\right)\right)^\frac{3}{4}}$$

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!

• 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. – Moti Jun 11 at 20:20
• you can not have negative values under the square roots... – Moti Jun 12 at 2:43
• (In less words, $(x+1)^2 \ge 1$ for any $x> 0$, and $\sqrt{x^2+2x} = \sqrt{(x+1)^2 - 1} \le \sqrt{(x+1)^2 } = x+1$.) – 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.
• 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 $y>1$. – Squirtle Jun 13 at 20:30