<s>[INST] 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.
 [/INST] 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,
$$a+b+c=3/2+x+y+z\le 3/2$$
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