To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
 A: Note that
$$
\left(\sqrt{x-1}+\sqrt{5-x}\right)^2=4+2\sqrt{4-(x-3)^2}
$$
Since $4-(x-3)^2\le4$ and $4+2\sqrt{4-(x-3)^2}\ge4$, we get that
$$
2\le\sqrt{x-1}+\sqrt{5-x}\le2\sqrt2
$$
A: Clearly $1\le x\le5$
If $y=\sqrt{x-1}+\sqrt{5-x}>0, y^2=4+2\sqrt{(x-1)(5-x)}$
Now $\sqrt{(x-1)(5-x)}\ge0$
and using AM-GM inequality, $\sqrt{(x-1)(5-x)}\le\dfrac{x-1+5-x}2=?$
A: Hint. Let $f(x)=\sqrt{x-1} + \sqrt{5-x}$ for $x\in[1,5]$.
i) Show that $f(x)=f(6-x)$ ($f$ is symmetric with respect to $x=3$).
ii) $f$ is strictly increasing in $(1,3)$, that is
$$f'(x)=\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{5-x}}>0$$
iii) $f([1,5])=[f(1),f(3)]$.
A: $$y=\sqrt{x-1}+\sqrt{5-x}\ge0$$ then
$$y^2=4+2\sqrt{x-1}\sqrt{5-x}\ge4$$ and 
$$(y^2-4)^2=4(x-1)(5-x)\ge0.$$
The positive range of the RHS is $[0,16]$. Then the range of $y^2-4$ is $[0,4]$ and that of $y$ is $[2,\sqrt8]$.
A: Clearly $1\le x\le5$
As $\dfrac{1+5}2=3, 1-3\le x-3\le5-3$
WLOG $x-3=2\cos2y,$  where $0\le2y\le\pi$
$\sqrt{x-1}=2\cos y$  and $\sqrt{5-x}=2\sin y$
Now $\sin y+\cos y=\sqrt2\sin\left(y+\dfrac\pi4\right)$ and $0\le y\le\dfrac\pi2$
$\implies\dfrac1{\sqrt2}\le\sin\left(y+\dfrac\pi4\right)\le1$
