<s>[INST] 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!
 [/INST] 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.
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