Without loss of generality, assume that $a_n$ is the largest number among all the $a_i$'s. Let's denote the arithmetic and geometric means of the first $n-1$ numbers by $A$ and $G$, respectively $$ A=\frac{a_1+a_2+\cdots+a_{n-1}}{n-1},\qquad G=\sqrt[n-1]{a_1a_2\cdots a_{n-1}} $$ Now the inequality that we want to prove in our induction step becomes $$ \sqrt[n]{G^{n-1}a_n} \le \frac{(n-1)A+a_n}{n} $$
Hello, I am seeking help in understanding the left-hand side of the second expression. I understand how they got the RHS, but the LHS confuses me.
The root in the expression for the previous $G$ will be $n-1$ so raising to $n-1$ will "undo" it.
$$\left(\sqrt[n-1]{\prod_i{a_i}}\right)^{n-1} = \prod_i a_i$$