Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$

Question

It's a charming problem :

Let $a,b,c>0$ such that $a+b+c=1$ then we have : $$\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$$

I know the identity :

Let $a,b,c>0$ such that $a+b+c=1$ then we have : $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=1.5$$

But I think it's not relevant here .

I try also majorization with the inequality :

Let $a\geq b\geq c>0$ such that $a+b+c=1$ then we have :

$$\exp\Big(\frac{-2}{3}\Big)a\geq \frac{a}{\exp(a+b)}$$

Second line of the majorization :

$$\exp\Big(\frac{-2}{3}\Big)^2ab\geq \frac{a}{\exp(a+b)}\frac{b}{\exp(b+c)}$$

Third line of the majorization :

$$\exp\Big(\frac{-2}{3}\Big)^3abc\geq \frac{a}{\exp(a+b)}\frac{b}{\exp(b+c)}\frac{c}{\exp(c+a)}$$

The lines are easy to check with the condition remains to apply Karamata's inequality and we are done . Unfortunately the second line fails .

My question : Have you a proof ?

Thanks a lot for sharing your time and knowledge .

Answer 1

The hint.

Rewrite our inequality in the following form. $$\sum_{cyc}be^a\leq\sqrt[3]e.$$

Show that for any $0<x<1$ the following inequality holds: $$e^x\leq\frac{9}{4}\left(e+\sqrt[3]e-4\right)x^3+\frac{3}{2}\left(10-e-5\sqrt[3]e\right)x^2+\frac{1}{4}\left(e+21\sqrt[3]e-28\right)x+1.$$ After this it's enough to prove that: $$\frac{9}{4}\left(e+\sqrt[3]e-4\right)\sum_{cyc}a^3b+\frac{3}{2}\left(10-e-5\sqrt[3]e\right)\sum_{cyc}a^2b+\frac{1}{4}\left(e+21\sqrt[3]e-28\right)\sum_{cyc}ab+1\leq\sqrt[3]e,$$ which after homogenization gives: $$\sqrt[3]e\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq$$ $$\geq 4\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)+e\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc).$$ Now, easy to see that $$\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq0,$$ $$\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)\geq0,$$ $$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0,$$ $$\sqrt[3]e>1.36$$ and $$e<2.72.$$ Thus, it's enough to prove that: $$1.36\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq$$ $$\geq 4\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)+2.72\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)$$ or $$\sum_{cyc}(18a^4-14a^3b+31a^3c-10a^2b^2-25a^2bc)\geq0$$ and since by Rearrangement $$\sum_{cyc}a^3c=abc\sum_{cyc}\left(a^2\cdot\frac{1}{b}\right)\geq abc\sum_{cyc}\left(a^2\cdot\frac{1}{a}\right)=\sum_{cyc}a^2bc,$$ it's enough to prove that $$\sum_{cyc}(18a^4-14a^3b+6a^3c-10a^2b^2)\geq0$$ or $$\sum_{cyc}(9a^4-7a^3b-5a^2b^2+3ab^3)\geq0$$ or $$\sum_{cyc}a(a-b)(9a^2+2ab-3b^2)\geq0$$ or $$\sum_{cyc}\left(a(a-b)(9a^2+2ab-3b^2)-2(a^4-b^4)\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(7a^2+7ab+2b^2)\geq0$$ and we are done!

Answer 2

Fact 1: $\mathrm{e}^x \le \frac{2}{3}x^2 + x + 1, \quad \forall x \le \frac{2}{3}$. (The proof is given at the end.)

The desired inequality is written as $$a\mathrm{e}^{2/3-a-b} + b\mathrm{e}^{2/3-b-c} + c\mathrm{e}^{2/3-c-a} \le 1.$$

Let $f(x) \triangleq \frac{2}{3}x^2 + x + 1$. By Fact 1, it suffices to prove that $$a f(2/3-a-b) + bf(2/3-b-c) + cf(2/3-c-a) \le 1.$$

With the substitutions $a = \frac{u}{u+v+w}, b=\frac{v}{u+v+w}, c = \frac{w}{u+v+w}$ for $u, v, w>0$, after clearing the denominators, it suffices to prove that $$7u^3-12u^2v+6u^2w+6uv^2-3uvw-12uw^2+7v^3-12v^2w+6vw^2+7w^3\ge 0$$ or $$(u^3+v^3+w^3-3uvw) + 6(u^3+v^3+w^3) - 12(u^2v+v^2w+w^2u) + 6(uv^2+vw^2+wu^2) \ge 0$$ which is obvious by using AM-GM (e.g.,$u^3 + uv^2 \ge 2u^2v$). We are done.

$\phantom{2}$

Proof of Fact 1: Let $g(x) = \ln (\frac{2}{3}x^2 + x + 1) - x$. We have $g'(x) = \frac{x(1-2x)}{2x^2+3x+3}$. Thus, $g(x)$ is strictly decreasing on $(-\infty, 0)$, strictly increasing on $(0, 1/2)$, strictly decreasing on $(1/2, \infty)$. Note also that $g(0) = 0$ and $g(\frac{2}{3}) > 0$. The desired result follows.