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!
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.