$ 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 
$a+b+c-abc<2$ ?
 A: Put $a=1-x,  b=1-y,c=1-z$.
Notice that $0<x,y,z<1$.
The inequality can be written as 
$3-(x+y+z)-(1-x)(1-y)(1-z) < 2$
After simplifying we get 
$xyz<xy+yz+zx$
This is trivial as $z<1 \implies xyz<xy \implies xyz<xy+yz+zx$
A: $0 \leq a,b,c \leq 1 \Rightarrow 0 \leq abc \leq 1$ 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 
$a+b+c -abc \leq \dfrac{2}{3}(a+b+c) \leq 2$
A: Define $f(a,b,c) = a+b+c-abc$. We calculate the gradient
$$
\nabla f = (1-bc,1-ac,1-ab).
$$
This is zero when $ab = ac = bc = 1$, thus $a=b=c=1$. This is clearly a maximum, so we have $f(a,b,c)< f(1,1,1) = 3 - 1 = 2$ for $(a,b,c)\neq(1,1,1)$.
A: $$\begin{align}
a+b+c-abc&=a+b+c(1-ab)\\
&\lt a+b+(1-ab)\\
&=1+a+b(1-a)\\
&\lt1+a+(1-a)=2
\end{align}$$
relying on $0\lt a,b,c\lt1$ for the steps $c(1-ab)\lt(1-ab)$ and $b(1-a)\lt(1-a)$.  It's reasonably clear that this is, in essence, a proof by induction of the general inequality
$$a_1+a_2+\cdots+a_n-(a_1a_2\cdots a_n)\lt n-1$$
if $0\lt a_1,a_2,\ldots,a_n\lt1$.
A: Treating $a$ and $b$ as constants and taking the derivative with respect to $c$ we get $1-ab$ which is positive, and so the value will be strictly less than when $c=1$. Plugging in $c=1$ gives a similar problem in 2 variables. 
A: Because $$a+b+c-abc=2-(1-a)(1-b)-(1-c)(1-ab)<2.$$
Done!
