 Solution

Let $d_K$ denote the field discriminant, and put $n=[K:\mathbb{Q}]\ge 2$. Let $r_1$ be the number of real field embeddings and $r_2$ the number of conjugate pairs of nonreal field embeddings, so $n=r_1+2r_2$. The Minkowski bound for ideal classes provides a nonzero integral ideal $A$ satisfyingRearranging gives the Minkowski lower bound for a number-field discriminantBecause $2r_2\le n$ and $\pi /4<1$, this is at leastApply the arithmetic-geometric mean inequality to $1,2,\dots ,n$:It follows thatwhere $n/(n+1)\ge 2/3$ for $n\ge 2$ was used. ThusThe source writes $|\Delta (K{)}^2|>1$. If $\Delta (K)$ denotes the embedding determinant, its square is $d_K$; if it denotes the field discriminant, the displayed result also implies the stated inequality. The conclusion therefore does not depend on that notational convention.