If $r_k=\frac{a_k}{S_n}-1$ for $S_n=\sum_{m=1}^n \lambda_{m}a_m$ and $1\le k\le n$, then $(r_k+1)^{\lambda_k}\le e^{\lambda_k r_k}$ for $(r_k)_{k=1}^n\in\mathbb{R_0^+}$ and for $(\lambda_k)_{k=1}^n\in\mathbb{R_0^+}$ since $x+1\le e^x$ for $x\in\mathbb{R_0^+}$. Thus $$\prod_{k=1}^n (r_k+1)^{\lambda_k}=\frac{\prod_{k=1}^n a_{k}^{\lambda_k}}{\left(\sum_{m=1}^n \lambda_m a_m \right)^{\sum_{k=1}^n \lambda_k}}$$ and $\prod_{k=1}^n e^{\lambda_k r_k}=e^{\sum_{k=1}^n \lambda_k r_k}=1.$ We can assume that $\sum_{k=1}^n \lambda_k =1$ without loss of generality, since $\lambda_k$ can be rescaled accordingly. Therefore $$\frac{\prod_{k=1}^n a_{k}^{\lambda _k}}{\sum_{m=1}^n \lambda_m a_m}=\frac{\prod_{k=1}^n a_{k}^{\lambda_k}}{\sum_{k=1}^n \lambda_k a_k}\le 1$$ which, using the substitution $\lambda_k =\frac{1}{n}$, gives $$\bbox[5px,border:2px solid red] {\sqrt[n]{\prod_{k=1}^n a_k}\le \frac{1}{n}\sum_{k=1}^n a_k.}$$
Now, which conditions must be met for $a_k$ in order to $$\left\lfloor \sqrt[n]{\prod_{k=1}^n a_k}\right\rceil =\left\lfloor\frac{1}{n}\sum_{k=1}^n a_k\right\rceil$$ where $\lfloor \rceil$ denotes the nearest integer function with the convention that half-integers are rounded to even numbers?
I thought of using the Fourier series, $$\lfloor x\rceil =x+\frac{1}{\pi}\sum_{j\gt 0}\frac{(-1)^j \sin 2\pi jx}{j}\quad x\notin\frac{\mathbb{Z}}{2},$$ though I don't know how would I get anything from that.