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 $a=b=c=\frac{2}{3}$.

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.
