China Mathematical Competition

The domain of $y$ is $x\in [0,13]$. We have $$\begin{array}{rl}y& =\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}\\ & =\sqrt{x+27}+\sqrt{13+2\sqrt{x(13-x)}}\\ & \ge \sqrt{27}+\sqrt{13}=3\sqrt{3}+\sqrt{13}.\end{array}$$ The equality holds when $x=0$. Therefore, the minimum of $y$ is $3\sqrt{3}+\sqrt{13}$.

On the other hand, by the Cauchy inequality we have $$\begin{array}{rl}{y}^2& =(\sqrt{x}+\sqrt{x+27}+\sqrt{13-x}{)}^2\\ & \le \left(\frac{1}{2}+1+\frac{1}{3}\right)\left[2x+(x+27)+3(13-x)\right]\\ & =121.\end{array}$$ The equality holds when $4x=9(13-x)=x+27$. It is so for $x=9$. Therefore, the maximum of $y$ is $11$.