Max:áp dụng BĐT bunhia ta có:$(\sqrt{x-1}+\sqrt{5-x})^2\leq(1^2+1^2)(x-1+5-x)$
$\Leftrightarrow y^2\leq 8\Rightarrow y\leq 2\sqrt{2}$
Min:
ta có$:y^2=(\sqrt{x-1}+\sqrt{5-x})^2=4+2\sqrt{(x-1)(5-x)}$
vì $:2\sqrt{(x-1)(x-5)}\geq 0\Rightarrow 2\sqrt{(x-1)(5-x)}+4\geq 4\Leftrightarrow y^2\geq 4\Leftrightarrow y\geq 2$