Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybody help me with this task? Especially the first condition seems very unusual to me. Thanks.

It is easy to see that the equality is achieved at a=b=c=1/2. Therefore, let's expand around that point, i.e. define $x,y,z$ by $a=x+1/2$, $b=y+1/2$, $c=z+1/2$. Then, the first equality becomes
$$1=2abc+a^2+b^2+c^2\\
\Longleftrightarrow 0=2xyz+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)$$
Next, Jensen's inequality gives you that the geometric mean is smaller than the arithmetic mean, i.e.
$$(x^2y^2z^2)^{1/3}\le \frac{x^2+y^2+z^2}{3}\\
\Longleftrightarrow |xyz|\le \frac{x^2+y^2+z^2}{3}\sqrt{\frac{x^2+y^2+z^2}{3}}\le\frac{x^2+y^2+z^2}{6}$$ 
where the last inequality follows from the fact that $-1/2<x,y,z<1/2$. Therefore, going back to the equality above
$$0\ge -\frac{x^2+y^2+z^2}{3}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)\ge\\
\ge -\frac{x^2+y^2+z^2}{2}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)=\\
=\frac{2xy+2xz+2zy+x^2+y^2+z^2}{2}+\frac{3}{2}(x+y+z)\\
\Longleftrightarrow x+y+z\le -\frac{(x+y+z)^2}{3}\le 0$$
Thus,
$$a+b+c=3/2+x+y+z\le 3/2$$
