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 $a=b=c=\frac{2}{3}$.
 A: Proof: By taking logarithm on both sides, it suffices to prove that 
$$a\ln a + b\ln b + c\ln c \ge \ln \tfrac{a^2b + b^2c + c^2a}{2}.$$
We will use the following bounds (their proof is not hard and thus omitted):
$$x\ln x \ge f(x) = \tfrac{2}{3}\ln \tfrac{2}{3} + (1 + \ln \tfrac{2}{3})(x- \tfrac{2}{3})
 + \tfrac{9}{20}(x-\tfrac{2}{3})^2, \quad \forall x\in (0, 2]$$
and
$$\ln\tfrac{4}{9} + \tfrac{9}{4}(y - \tfrac49) \ge \ln y, \quad \forall y > 0.$$
With the bounds above, it suffices to prove that
$$f(a) + f(b) + f(c) \ge \ln\tfrac{4}{9} + \tfrac{9}{4}(\tfrac{a^2b + b^2c + c^2a}{2} - \tfrac49)$$
which is simplified to (by using $a+b+c=2$)
$$18a^2+18b^2+18c^2+16 - 45a^2b - 45b^2c - 45c^2a \ge 0.$$
After homogenization, it suffices to prove that
$$(18a^2+18b^2+18c^2)\tfrac{a+b+c}{2}+16(\tfrac{a+b+c}{2})^3- 45a^2b - 45b^2c - 45c^2a \ge 0.$$
The Buffalo Way works. WLOG, assume that $c = \min(a, b, c)$.
Let $b = c + s, \ a = c + t$ for $s, t\ge 0$.
It suffices to prove that
$$(18s^2-18st+18t^2)c +11s^3+15s^2t-30st^2+11t^3\ge 0.$$
It is not hard to prove that $11s^3+15s^2t-30st^2+11t^3\ge 0$ for $s, t\ge 0$. We are done.
A: Hints :In fact the key of this inequality is the following lemma :

$$2(a+\varepsilon)^{a+\varepsilon}b^b(c-\varepsilon)^{c-\varepsilon}\geq 2a^a b^b c^c \geq a^2b+b^2c+c^2a\geq (a+\varepsilon)^2b+b^2(c-\varepsilon)+(a+\varepsilon)(c-\varepsilon)^2$$
  With $0<\varepsilon<c$ and $a+\varepsilon\geq c-\varepsilon $

Proof :
We have to prove (for the first part)
$$2(a+\varepsilon)^{a+\varepsilon}b^b(c-\varepsilon)^{c-\varepsilon}\geq 2a^a b^b c^c$$
Wich is equivalent to :
$$(a+\varepsilon)ln(a+\varepsilon)+(c-\varepsilon)ln(c-\varepsilon)\geq aln(a)+cln(c)$$
Now we use Niculescu's inequality to solve this  :

Let $a,b,c,d$  be real positive numbers such that$$ a\geq c , b\leq d ,c\geq d $$
  And $f$ be a convex function we have :

$$0.5(f(a)+f(b))-f((a+b)0.5)\geq 0.5(f(c)+f(d))-f(0.5(c+d))$$
It's easy to remark that we have :
$$ a+\varepsilon\geq a , c\geq c-\varepsilon ,a+\varepsilon\geq c-\varepsilon $$
And that $xln(x)$ is convex . 
The second part is easy to show and now you can build the inequality !
A: 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.
