$ 0< a,b,c <1\implies a+b+c-abc<2$

Question

If $a,b,c$ are positive real numbers , all being less than $1$ , then how to prove that

$a+b+c-abc<2$ ?

Answer 1

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 $z<1 \implies xyz<xy \implies xyz<xy+yz+zx$

Answer 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

$a+b+c -abc \leq \dfrac{2}{3}(a+b+c) \leq 2$

Answer 3

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 $(a,b,c)\neq(1,1,1)$.

Answer 4

$$\begin{align} a+b+c-abc&=a+b+c(1-ab)\\ &\lt a+b+(1-ab)\\ &=1+a+b(1-a)\\ &\lt1+a+(1-a)=2 \end{align}$$

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 $0\lt a_1,a_2,\ldots,a_n\lt1$.

Answer 5

Because $$a+b+c-abc=2-(1-a)(1-b)-(1-c)(1-ab)<2.$$ Done!

Answer 6

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.