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

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!