# A conjecture about power sum : $e^{ab}+e^{bc}+e^{ca}\geq 3e^{\sqrt{abc}}$ and $a+b+c=3$

Hello I have this to propose :

Let $$a,b,c$$ be real positive numbers such that $$a+b+c=3$$ then we have : $$e^{ab}+e^{bc}+e^{ca}\geq 3e^{\sqrt{abc}}$$

For a generalization I have this conjecture :

Let $$a_i$$ be $$n$$ real positive numbers such that $$\sum_{i=1}^{n}a_i=n$$ then we have (with $$a_{n+1}=a_1$$): $$\sum_{i=1}^{n}e^{a_ia_{i+1}}\geq ne^{\Big(\prod_{i=1}^{n}a_i\Big)^{\frac{1}{n-1}}}$$

In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .

By Jensen $$e^{ab}+e^{ac}+e^{bc}\geq3e^{\frac{ab+ac+bc}{3}}\geq3e^{\sqrt{abc}}$$ because the last inequality it's $$ab+ac+bc\geq\sqrt{3(a+b+c)abc}$$ or after squaring of the both sides $$\sum_{cyc}c^2(a-b)^2\geq0.$$ The second inequality is wrong.
Try $$n=4$$, $$a_1=a_3=\frac{1}{4}$$ and $$a_2=a_4=\frac{7}{4}.$$
For the first, recall $$x\mapsto e^x$$ is convex. Thus, by Jensen's inequality, $$f(ab)+f(bc)+f(ca)\geq 3f(\frac{ab+bc+ca}{3})=3\exp(\frac{ab+bc+ca}{3}),$$ where $$f(x)=e^x$$. Now, observe that, $$(ab+bc+ca)^2\geq 3abc(a+b+c)=9abc$$. Hence, $$\frac{ab+bc+ca}{3}\geq \sqrt{abc}$$. and therefore, using the fact that $$f(\cdot)$$ is increasing, we conclude, $$e^{ab}+e^{bc}+e^{ca}\geq 3f(\frac{ab+bc+ca}{3})\geq 3e^{\sqrt{abc}}.$$ For the second one, I am very confident that this approach should transfer without much hassle.