What remains is to show that \(\llcorner n/3\lrcorner\) or fewer guards can be placed on vertices so that every triangle is in the view of a guard. Let \(b\), \(r\) and \(w\) be the number of vertices coloured blue, red and white respectively. Let \(b\) be minimal in \(\{b, r, w\}\). Then \(b \le r\) and \(b \le w\). Then \(2b \le r + w\). So \(3b \le b + r + w = n\). So \(b \le n/3\) and so \(b \le \llcorner n/3\lrcorner\). Place a guard on each blue vertex. Done.