SELECTION EXAMINATION

a. Since $x\ge n^2$, we have $$n\sqrt{x-n^2}\le \frac{x}{2}\iff 2n\sqrt{x-n^2}\le x\iff 4n^2(x-n^2)\le x^2\iff (x-2n^2{)}^2\ge 0,$$ which is valid. Equality holds if and only if $x=2n^2$.

Alternatively, for every $x\ge n^2$, it is enough to prove that $$n\sqrt{x-n^2}-\frac{x}{2}\le 0\iff 2n\sqrt{x-n^2}-x\le 0\iff \frac{(2n\sqrt{x-n^2}-x)(2n\sqrt{x-n^2}+x)}{2n\sqrt{x-n^2}+x}\le 0$$ $$\iff \frac{4n^2(x-n^2)-x^2}{2n\sqrt{x-n^2}+x}\le 0\iff \frac{-(x-2n^2{)}^2}{2n\sqrt{x-n^2}+x}\le 0,$$ which is valid. Equality holds for $x=2n^2$.

b. The given inequality can be written in the form $$(2\sqrt{x-1}-x)+(4\sqrt{y-4}-y)+(6\sqrt{z-9}-z)=0,(1)$$ for $x\ge 1$, $y\ge 4$ and $z\ge 9$.

Using (a) for $n=1,2,\dots ,3$, we get $$2\sqrt{x-1}-x\le 0,4\sqrt{y-4}-y\le 0\text{and}6\sqrt{z-9}-z\le 0,$$ And therefore (1) is possible to be valid, only for $$2\sqrt{x-1}-x=0,4\sqrt{y-4}-y=0\text{and}6\sqrt{z-9}-z=0\iff x=2,y=8,z=18.$$