China Mathematical Competition (Shaanxi)

By Cauchy's inequality, we have $$\begin{array}{rl}& 2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}\\ & =\sqrt{x+1}+\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}\\ & \le \sqrt{[(x+1)+(x+1)+(2x-3)+(15-3x)](1^2+1^2+1^2+1^2)}\\ & =2\sqrt{x+14}\le 2\sqrt{19},\end{array}$$ and the equality holds if and only if $\sqrt{x+1}=\sqrt{2x-3}=\sqrt{15-3x}$ and $x=5$. But this is impossible. So $2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}$.