Prove the maximum of the function $\det x$ restrained to the sphere $\sum x_{ij}^2=n$ is $1$

The sum $\sum x_{ij}^2 = \operatorname{tr} x^t x$, so the eigenvalues $\lambda_1, \dots, \lambda_n$ of the diagonalizable matrix $x^t x$ satisfy $\lambda_1 + \cdots + \lambda_n = n$. Furthermore, each $\lambda_i \geq 0$, since any eigenvector $v_i$ belonging to $\lambda_i$ has $$\lambda \langle{v_i, v_i}\rangle = \langle{x^t x v_i, v_i}\rangle = \langle{x v_i, x v_i}\rangle \geq 0.$$ Now show that $(\det x)^2 = \det x^t x = \lambda_1 \cdots \lambda_n$ is maximized, subject to the constraints above on the $\lambda_i$, when all $\lambda_i = 1$.