Let $a_i, \;\; i =1, \dots, n$ be positive real numbers. Prove that:
$$\sqrt[n]{\prod_{i=1}^{n}\prod_{j=1}^{n}\left ( 1+\frac{a_i}{a_j} \right )}\geq 2^n, \;\; n \in \mathbb{N}$$
Solution
We are invoking the very useful inequality
$$1+\frac{a}{b}\geq 2 \sqrt{\frac{a}{b}} \tag{1} \label{*}$$
Now we are applying \eqref{*} $n^2$ times hence
$$\sqrt[n]{\prod_{i=1}^{n}\prod_{j=1}^{n}\left ( 1+\frac{a_i}{a_j} \right )} \geq \sqrt[n]{2^{n^2} \prod_{i=1}^{n}\prod_{j=1}^{n}\sqrt{\frac{a_i}{a_j}}} \geq 2^n$$
$$\sqrt[n]{\prod_{i=1}^{n}\prod_{j=1}^{n}\left ( 1+\frac{a_i}{a_j} \right )}\geq 2^n, \;\; n \in \mathbb{N}$$
Solution
We are invoking the very useful inequality
$$1+\frac{a}{b}\geq 2 \sqrt{\frac{a}{b}} \tag{1} \label{*}$$
Now we are applying \eqref{*} $n^2$ times hence
$$\sqrt[n]{\prod_{i=1}^{n}\prod_{j=1}^{n}\left ( 1+\frac{a_i}{a_j} \right )} \geq \sqrt[n]{2^{n^2} \prod_{i=1}^{n}\prod_{j=1}^{n}\sqrt{\frac{a_i}{a_j}}} \geq 2^n$$
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