Solution
= Solution
{c}
If every $\|v_i\|=1$, then each diagonal entry of the <Gram matrix> is $A_{ii}=\langle v_i,v_i\rangle=1$, so its <matrix trace> is $n$. Using <4/b/iii> and the permitted <arithmetic-geometric mean inequality>,
$$
0\leq\det A=\prod_{i=1}^n\lambda_i
\leq\left(\frac{\sum_{i=1}^n\lambda_i}{n}\right)^n
=\left(\frac{\operatorname{tr}A}{n}\right)^n
=\boxed{1}.
$$
The bound is attained by any <orthonormal> family, whose <Gram matrix> is the <identity matrix>.