Let $a_i>0$ be $n$ numbers such that $\prod_{i=1}^{n}a_i\leq 1$ then we have: $$\sum_{i=1}^{n}a_i\geq n \left(\prod_{i=1}^{n}a_i \right)^{\large\left({n+\sum_{i=1}^{n}a_i-n(\prod_{i=1}^{n}a_i)^{1/n}}\right)^{-1}}.$$
A possible way to generalize this is the following conjecture:
Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $\prod_{i=1}^{n}f(a_i)\leq 1$ then we have: $$\sum_{i=1}^{n}f(a_i)\geq n \left( \prod_{i=1}^{n}f(a_i) \right)^{\large \left(n+\sum_{i=1}^{n}f(a_i)-nf\left(\sum_{i=1}^{n}a_i/n\right)\right)^{-1}}.$$
Any hints would be appreciable to solve this. Thanks in advance.
For the first inequality, write $A=\tfrac1n \sum a_i$ and $G=\prod a_i^{1/n}.$ We need to prove $A^{1+A-G}\geq G.$ But $$A^{A-G}=(A^A)^{(A-G)/A}\geq e^{-(A-G)/A}\geq 1-(A-G)/A=G/A$$ where the middle inequality uses $A^A\geq e^{-e^{-1}}\geq e^{-1}.$
The generalization is false, if my calculations are correct. Take \begin{align*} n&=4\\ a_1&=\tfrac12\\ a_i&=1\qquad(2\leq i\leq 4)\\ f(x)&=\begin{cases}x/e^2& (x\leq 7/8)\\ 7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (x\geq 7/8) \end{cases}\\ f(a_1)&=1/2e^2\\ f(a_i)&=1/e\qquad(2\leq i\leq 4)\\ A&=\tfrac1n\sum f(a_i)\approx0.29283\\ G&=\prod f(a_i)^{1/n}=(1/2e)^{1/4}/e\approx0.24092\\ F&=f(\sum a_i/n)=f(7/8)=7/8e^2\approx 0.11842. \end{align*} Then $0.80718\approx A^{A-F}<G/A\approx 0.82274,$ but the desired inequality is equivalent to $A^{A-F}\geq G/A.$