I see on a french forum the following inequality :
Let $a_i>0$ be $n$ real numbers such that $\sum_{i=1}^{n}a_i\ln(a_i)=0$ then we have :
$$n\Big(\prod_{i=1}^{n}a_i\Big)^{\frac{1}{n}}\leq n\sqrt{\frac{n}{\sum_{i=1}^{n}\frac{1}{a_i}}}\leq \sum_{i=1}^{n}a_i$$
For $n=2$ :
The LHS is equivalent to :
$$a_1+a_2\leq 2$$
Wich is true with the condition
The RHS is equivalent to :
$$\Big(\frac{a_1+a_2}{2}\Big)^3\geq a_1a_2$$
Wich is easy to prove .
The general case is more delicate and I don't know how to deal with it .
Furthermore the inequality is non-homogeneous and we cannot use Jensen's inequality .
Finally I can't find the page on the french forum where there is inequality .
If you have a hint it would be cool.
Thanks a lot for sharing your time and knowledge .
Ps : I find beautiful this refinement because we have the most important mean in mathematics.
Best Answer
The conjecture $$ n\Big(\prod_{i=1}^{n}a_i\Big)^{\frac{1}{n}}\leq n\sqrt{\frac{n}{\sum_{i=1}^{n}\frac{1}{a_i}}}\leq \sum_{i=1}^{n}a_i $$ holds for $n=2$. For higher $n$, it does not hold, and there are counterexamples where either side of the double inequality is violated.
Here are samples which meet the condition $\sum_{i=1}^{n}a_i\ln(a_i)=0$. Let $n=3$.
The left inequality is violated e.g. for
$a_1 = 0.6, a_2 = 1.3, a_3 = 0.006961$, where upon inserting these numbers the proposed double inequality reads
$0.527282 \le 0.429898 \le 1.906961$.
The right inequality is violated e.g. for
$a_1 = 0.7, a_2 = 1.7, a_3 = 0.275730$, where upon inserting these numbers the proposed double inequality reads
$2.438257 \le 2.750378 \le 2.675730$.
Both samples violate the conjecture. $\qquad \Box$