$ 0< a,b,c <1\implies a+b+c-abc<2$ If $a,b,c$ are positive real numbers , all being less than $1$ , then how to prove that 
$a+b+c-abc<2$ ?

$0 \leq a,b,c \leq 1 \Rightarrow 0 \leq abc \leq 1$ and $0 \leq a+b+c \leq 3$. 
It follows that $\sqrt[3]{abc} \geq abc$ and by the AM-GM inequality we know 
$\dfrac{a+b+c}{3} \geq \sqrt[3]{abc} \geq abc$.
Finally 
$a+b+c -abc \leq \dfrac{2}{3}(a+b+c) \leq 2$
