Given $a,b,c$ are non-negative number satisfy $a+b+c=1$. Find the value of maximize $$B=ab+bc+ca-3abc$$
\begin{align}B&=(a+b+c)(ab+bc+ca)-3abc\\ &=a^2(b+c)+b^2(a+c)+c^2(a+b)\\ &=a^2(1-a)+b^2(1-b)+c^2(1-c)\end{align}
If $a=b=c=\frac{1}{3} \Rightarrow B_{max}=\frac{2}{9} $
We need $\sum_{cyc} a^2-\sum_{cyc}a^3 \le \frac{2}{9}$
$\Rightarrow (\sum_{cyc}a^2-\frac{1}{3})(\sum_{cyc}a^2-\frac{2}{3})\ge 0$
The last inequality seem wrong because It is possibly $a^2+b^2+c^2\le \frac{2}{3}$, help me !
Notice that
\begin{align*} \frac{1}{4}-B &= \frac{1}{4}(a+b+c)^3 - \sum_{perm}a^2 b \\ &= \frac{1}{4}\sum_{cyc} a^3 - \frac{1}{4} \sum_{perm} a^2b + \frac{6}{4}abc \\ &= \frac{1}{4}\left( \sum_{cyc} a(a-b)(a-c) \right) + \frac{3}{4}abc. \end{align*}
The first term is $\geq 0$ by the Schur's inequality, and the second term is obviously $\geq 0$. So the entire quantity is $\geq 0$ and we have $B \leq \frac{1}{4}$.
Finally, plugging $(a,b,c) = (\frac{1}{2}, \frac{1}{2}, 0)$ gives $B = \frac{1}{4}$ and therefore $B_{\max} = \frac{1}{4}$.
Here is an unnecessarily and excessively complicated solution:
Consider the hyperplane $H$ in $\Bbb{R}^3$ defined by the equation $a+b+c = 1$. We identify $H$ with $\Bbb{R}^2$. Then the 2D-Laplaican $\Delta_H$ on $H$ can be expressed in terms of the coordinates of $\Bbb{R}^3$ as:
$$ \Delta_H f(a,b,c) = \frac{2}{3}\left( \frac{\partial^2f}{\partial a^2} + \frac{\partial^2f}{\partial b^2} + \frac{\partial^2f}{\partial c^2} - \frac{\partial^2f}{\partial a \partial b} - \frac{\partial^2f}{\partial b \partial c} - \frac{\partial^2f}{\partial c \partial a} \right). $$
Now it is just an easy computation to check that
$$ \Delta_H B = 2(a+b+c-1) = 0. $$
Thus $B$ is harmonic on $H$. Let $\Omega = \{(a,b,c) \in H : a,b,c \geq 0 \}$. Then by the maximum principle, the maximum of $B$ is achieved on the boundary $\partial \Omega$. First, along the boundary with $c = 0$ we have
$$ B(a, 1-a, 0) = a(1-a) \leq \frac{1}{4} \quad \text{with equality iff} \quad a = \frac{1}{2}. $$
By the obvious symmetry, similar observations are true along other boundary segments. Therefore $B_{\max} = \frac{1}{4}$ and this is achieved exactly at 3 points $(\frac{1}{2},\frac{1}{2},0)$, $(\frac{1}{2},0,\frac{1}{2})$, $(0,\frac{1}{2},\frac{1}{2})$.