There's Only One Equation?

Relevant wiki: Algebraic Manipulation - Identities - Basic

Let $\begin{cases} u^2 = x+1 & \implies x = u^2 - 1 \\ v^2 = y & \implies y = v^2 \\ w^2 = z-4 & \implies z = w^2 + 4 \end{cases}$

$\begin{aligned} \implies \sqrt{x+1} + \sqrt{y} + \sqrt{z-4} & = \frac {x+y+z}2 \\ 2(\sqrt{x+1} + \sqrt{y} + \sqrt{z-4}) & = x+y+z \\ 2(u+v+w) & = u^2-1+ v^2 + w^2+4 \quad \quad \small \color{#3D99F6}{\text{Rearranging}} \\ u^2 - 2u+1 + v^2 - 2v+1 + w^2 - 2w+1 & = 0 \\ (u-1)^2 + (v-1)^2 + (w-1)^2 & = 0 \quad \quad \small \color{#3D99F6}{\text{As LHS }\ge 0 \text{; equality iff }u=v=w=1} \end{aligned}$

$\implies \begin{cases} x = u^2 -1 & = 0 \\ y = v^2 & = 1 \\ z = w^2+4 & = 5 \end{cases} \implies x+y+z = \boxed{6}$

Relevant wiki: Algebraic Manipulation Problem Solving - Basic

Rearrange equation to write:-

$\small{\left[\{(x+\color{#D61F06}{1})-2\sqrt{x+1}+\color{#D61F06}{1})\}+\{y-2\sqrt y+\color{#D61F06}{1}\}+\{(z\color{#D61F06}{-4})-2\sqrt{z-4}+\color{#D61F06}{1}\}\right]=0}$ $\small{\implies \left[\sqrt{x+1}-1\right]^2+\left[\sqrt y-1\right]^2+\left[\sqrt{z-4}-1\right]^2=0}$

$\small{\implies \sqrt{ x+1}-1=\sqrt y-1=\sqrt{z-4}-1=0}$ $\small{\implies \sqrt{x+1}=\sqrt y=\sqrt{z-4}=1}$ $\implies x=0,y=1,z=5$

$\huge \therefore ~0+1+5=\color{#007fff}{\boxed 6}$

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

It is clear that $x+1$ , $y$ and $z-4$ are non-negative,

Then applying AM-GM on $(x+1)$ and 1 we have $\frac{x+2}{2}\geq\sqrt{x+1}$

Now apply AM-GM on $y$ and 1 to get $\frac{y+1}{2}\geq\sqrt{y}$

And finally AM-GM on $(z-4)$ and 1 to get $\frac{z-3}{2}\geq\sqrt{z-4}$

Adding all three we have $\sqrt{x+1}+\sqrt{y}+\sqrt{z-4}\leq\frac{x+y+z}{2}$

The equation asks us for strict equality which for AM-GM only happens when all the terms are equal so,

$x+1=1 \implies x=0$

$y=1$

$z-4=1 \implies z=5$

Thus $x+y+z=6$