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 .

Thanks in advance .

  • $\begingroup$ @MartinR Maybe OP is trying to create a contest-math problem. $\endgroup$ – yurnero Feb 1 at 13:42

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.


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