$ĐK : x \geq 2$Vì cả hai vế cùng lớn hơn 0
Bình phương hai vế, ta có
$x+1+4x-4+4\sqrt{x+1}{x-2} \leq 5x+1$
$\Leftrightarrow 4\sqrt{x^2-x-2} \leq 4$
$\Leftrightarrow 0 \leq x^2-x-2 \leq 1$
$\Leftrightarrow \left\{ \begin{array}{l} x^2-x-2 \geq 0\\ x^2-x-2\leq 1\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x\leq -1, x\geq 2\\ -1\leq x\leq 2\end{array} \right.$
$\Rightarrow x=-1;2$