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