Question
If $a_j\gt 0$, $j=1,\ldots,m$ and $x_i\ge 0$ $i=1,\ldots,n.\;$ then prove that
\begin{align} \sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}&\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}\\ &\ge\prod_{i=1}^{n}\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}. \end{align}
What do i do so far
I start with the left inequality
$\sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}$
For any positive Integers
$\sqrt[k]{a_1 \cdot a_2 \cdot \ldots \cdot a_k} \ge \frac{a_1 + a_2 + \ldots + a_k}{k}$
Let K=M
$\sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}$
By AM-GM INEQUALITY
$\sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]} \ge \frac{a_1 + a_2 + \ldots + a_m + \frac{1}{n}\sum_{i=1}^{n}x_i}{m}$
Right hand inequality $(\frac{1}{n}\sum_{i=1}^{n}(a_1+x_i + a_2+x_i + \ldots + a_m+x_i))$ use properties of summation $(\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{m}(a_j+x_i))$
$\frac{a_1 + a_2 + \ldots + a_m + \frac{1}{n}\sum_{i=1}^{n}x_i}{m} = \frac{1}{n}\sum_{i=1}^{n}\frac{1}{m}\sum_{j=1}^{m}(a_j+x_i)$
The right inequality it's just AM-GM: $$\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)} \ge\sqrt[n]{\prod_{i=1}^n\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}}=\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}.$$
The left inequality it's just Holder: $$\sqrt[m]{\prod_{j=1}^{m}\left(a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right)}=\frac{1}{n}\sqrt[m]{\prod_{j=1}^{m}\sum_{i=1}^n(a_j+x_i)}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)}.$$