min of $c$ in such that $\frac{\sum^n_{i=1}a_i}{n\sqrt[n]{\prod^n_{i=1}a_i}}\leq \left(\frac{\sqrt{a/b}+\sqrt{b/a}}2\right)^c$ for all $a_i\in [a,b]$.

Question

Positive integer $n\geq 2$ is given. All $a_i\in[a,b]$($0<a<b$). What's the minimum value of $c$ such that $$\tag{1}\frac{\sum^n_{i=1}a_i}{n\sqrt[n]{\prod^n_{i=1}a_i}}\leq \left(\frac{\sqrt{\frac{a}b}+\sqrt{\frac{b}a}}2\right)^c$$

We can show that $a_i=a\ {\rm or}\ b$ when LHS reach the max value, by method of adjustment:

Suppose that $t=\sqrt[n]{a_1}$, then $$f(t)=\frac{\sum^n_{i=1}a_i}{\sqrt[n]{\prod^n_{i=1}a_i}}=\frac{t^{n-1}}{\sqrt[n]{a_2a_3\cdots a_n}}+\frac{a_2+a_3+\cdots+a_n}{t\sqrt[n]{a_2a_3\cdots a_n}}$$ is concave since $f''(t)>0$. Then when the LHS reach the maximum value, $a_1$ should equal to $a$ or $b$. The same to the other $a_i$. So we can suppose that $a_i=a\ {\rm or}\ b$ without lost of generality.

Also, it can be proved by weighted AM-GM inequality that $c=2-\frac2n$ satisfies the inequality (1).

The question is, what's the minimum value of $c$?(Maybe $c$ is a function of $n$.)