Macedonian Mathematical Olympiad

Let $a_i^2={\tan }^2{x}_i$, $x_i\in [0,\frac{\pi }{2}]$, $i=1,2,\dots ,n$. Then ${\sum }_{i=1}^n{\cos }^2{x}_i=1$. From the inequality between the arithmetical and the geometrical mean it follows that $${\sin }^2{x}_i=1-{\cos }^2{x}_i\ge (n-1){\left(\prod _{j=1,j\ne i}^n\cos x_j\right)}^{2/(n-1)},i=1,2,\dots ,n.$$ By multiplying the $n$ inequalities above, we get ${\prod }_{i=1}^n{\sin }^2{x}_i\ge (n-1{)}^n{\prod }_{i=1}^n{\cos }^2{x}_i$. The last inequality is equivalent to the inequality $$\prod _{i=1}^n\tan x_i\ge (n-1{)}^{n/2}.$$ Finally, ${\prod }_{i=1}^n{a}_i=({\prod }_{i=1}^n\tan x_i{)}^{1/2}\ge (n-1{)}^{n/4}$, which was to be proven.