Inspired by nam le - Part 3

Let $a = 16x+9, b = 16y+9, c = 16z+9.$ Then $a,b,c \ge 0,$ and $a+b+c = 16 \cdot 2018+27.$ Call this number $N.$ We want the max and min of the expression $\sqrt{a}+\sqrt{b}+\sqrt{c}.$

Note that for $m,n \ge 0,$ we have $\sqrt{m}+\sqrt{n} \ge \sqrt{m+n}$ (proof: square both sides). So $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge \sqrt{a} + \sqrt{b+c} \ge \sqrt{a+b+c},$ so the minimum is clearly $\sqrt{N},$ achieved when $a=b=0$ and $c=N.$

Also note that for $m,n,p,q \ge 0,$ we have $m\sqrt{p} + n\sqrt{q} \le \sqrt{m+n}\sqrt{mp+nq}.$ Proof: the right side squared minus the left side squared is $mn(p+q)-2mn\sqrt{pq} = mn(\sqrt{p}-\sqrt{q})^2.$ We'll use this twice: in particular, $\sqrt{p}+\sqrt{q} \le \sqrt{2}\sqrt{p+q}$ and $\sqrt{p}+2\sqrt{q} \le \sqrt{3}\sqrt{p+2q}.$

So, $\sqrt{a}+\sqrt{b}+\sqrt{c} \le \sqrt{a} + \sqrt{2}\sqrt{b+c} = \sqrt{a} + 2\sqrt{\frac{b+c}2} \le \sqrt{3} \sqrt{a+b+c}.$ This implies that the maximum value of $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is $\sqrt{3N},$ which is indeed achieved when $a=b=c=N/3.$

So the ratio is $\frac{\sqrt{3N}}{\sqrt{N}} = \sqrt{3} \approx \fbox{1.7321}.$

It doesn't really matter what $N$ is here, so there was nothing important about $2018$ in the original problem.