# 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 $a_i\in [a,b]$.

Positive integer $$n\geq 2$$ is given. All $$a_i\in[a,b]$$($$0). What's the minimum value of $$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 $$n$$.)

• @TaD Denote the maximum of LHS by $f(a, b, n)$. Do you want to find the minimum of $c$ such that $f(a, b, n) \le RHS$ for all $b > a > 0$ (as a function of $n$)? Aug 22, 2022 at 23:04
• @River Li, I think you're right. (If my above derivation is right.)