But it is useful because it provides the idea for a proof of the conclusion—one of the major benefits of visual thinking in mathematics. In brief: for each equation \((x\cdot y = k\); \(x + y = 2\sqrt{k})\) if \(x = y\), their common value is \(\sqrt{k}\). So the functions expressed by those equations meet at the diagonal. To show that, for a fixed positive \(x\)-value, the \(y\)-values of \(x\cdot y = k\) never fall below the \(y\)-values of \(x + y = 2\sqrt{k}\), it suffices to show that \(2\sqrt{k} - x \le k/x\). As a geometric mean is less than or equal to the corresponding arithmetic mean, \(\sqrt{[x \cdot (k/x)]} \le [x + (k/x)]/2\).  So \(2\sqrt{k} \le x + (k/x)\). So \(2\sqrt{k} - x \le k/x\).