<s>[INST] For $a+b+c=2$ prove that $2a^ab^bc^c\geq a^2b+b^2c+c^2a$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove that:
$$2a^ab^bc^c\geq a^2b+b^2c+c^2a$$
I tried convexity, but without success:
We need to prove that
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln\sum_{cyc}a^2b$$ and since $f(x)=x\ln{x}$ is a convex function, by Jensen we obtain:
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln2+3\cdot\frac{2}{3}\ln\frac{2}{3}=\ln\frac{8}{9}.$$ 
Thus, we need to prove that
$$\frac{8}{9}\geq\sum_{cyc}a^2b$$ or
$$(a+b+c)^3\geq9(a^2b+b^2c+c^2a),$$ which is wrong for $c\rightarrow0^+$.
The equality occurs for $a=b=c=\frac{2}{3}$.
 [/INST] Remark: My first proof does not work for the stronger inequality below.
Let us prove a stronger inequality:
$$2a^a b^b c^c \ge \frac{4}{27}(a + b + c)^3 - abc.$$
(Note: It is well-known that $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$.)
Letting $x = \frac32 a, y = \frac32 b, z = \frac32 c$, it suffices to prove that, for all $x, y, z > 0$ with $x + y + z = 3$,
$$3 (x^x y^y z^z)^{2/3} \ge \frac{4}{27}(x+y+z)^3 - xyz. $$
(Note: We have $a^a b^b c^c
= (2/3)^{2(x+y+z)/3}(x^xy^yz^x)^{2/3} = \frac49 (x^xy^yz^x)^{2/3} $.)
Using $\mathrm{e}^u \ge 1 + u$, we have
$$(x^x y^y z^z)^{2/3}
= \mathrm{e}^{\frac23(x\ln x + y\ln y + z\ln z)}
\ge 1 + \frac23(x\ln x + y\ln y + z\ln z).$$
It suffices to prove that
$$3 + 2x\ln x + 2y\ln y + 2z\ln z \ge \frac{4}{27}(x+y+z)^3 - xyz.$$
Fact 1: It holds that
$u\ln u \ge u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3
= -\frac13 - \frac12 u + u^2 - \frac16 u^3$
for all $u > 0$.
(The proof is given at the end. Note: The RHS is $3$-th order Taylor approximation of $u\ln u$ around $u = 1$.)
Using Fact 1, it suffices to prove that
\begin{align*}
 &3 + \left(-\frac23 - x + 2x^2 - \frac13 x^3\right)
 + \left(-\frac23 - y + 2y^2 - \frac13 y^3\right)\\
 &\quad + \left(-\frac23 - z + 2z^2 - \frac13 z^3\right)\\
 &\ge \frac{4}{27}(x+y+z)^3 - xyz
\end{align*}
or (using $x + y + z = 3$)
$$3 - xy - yz - zx \ge 0$$
which is true.
We are done.

Proof of Fact 1:
It suffices to prove that
$$\ln u \ge \frac{1}{u}\left(u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3\right).$$
Let $F(u) := \mathrm{LHS} - \mathrm{RHS}$. We have
$$F'(u) = \frac{(u-1)^3}{3u^2}.$$
We have $F'(u) < 0$ on $(0, 1)$, and $F'(u) > 0$ on $(0, \infty)$, and $F'(1) = 0$.
Thus, $F(u) \ge F(1) = 0$ for all $u > 0$.
We are done.
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