How about this solution for the inequalities question?
$Since $ (x-1)^2 \geq 0, $ then$ \\ (x-1)(x-1)\geq 0 \\ (x-1)(\sqrt x - 1)(\sqrt x +1) \geq 0 $ as $ x \geq 0 \\ $But, $ \sqrt x + 1 >0 $ for all $ x \geq 0 \\ $then $ (x-1)(\sqrt x-1)\geq 0 \\ $so upon expanding, $ x\sqrt x + 1 \geq x+\sqrt x