Hello I can't do this exercise please help me
Let a, b, c be positive real numbers such that abc=1. Prove that
(a-1+1/b)(b-1+1/c)(c-1+1/a)lessthan or eqal1
$\displaystyle (a - 1 + \frac{1}{b})(b - 1 + \frac{1}{c})(c - 1 + \frac{1}{a}) $Originally Posted by luckyvan
$\displaystyle (\frac{{ab - b + 1}}{b})(\frac{{bc - c + 1}}{c})(\frac{{ac - a + 1}}{a}) $
$\displaystyle \frac{{(ab - b + 1)(bc - c + 1)(ac - a + 1)}}{{abc}} $
$\displaystyle \frac{{(ab^2 c - abc + ab - b^2 c + bc - b + bc - c + 1)(ac - a + 1)}}{{abc}} $
$\displaystyle \frac{{(ab - b^2 c + 2bc - c)(ac - a + 1)}}{{abc}} $
$\displaystyle a + b + c - a^2 b - b^2 c - ac^2 + ab + bc + ac - 2 $
$\displaystyle ab(1 - a) + bc(1 - b) + ac(1 - c) + a + b + c - 2 $
$\displaystyle a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1) - 2 $
Since a,b,c are all positive then
$\displaystyle a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1)$ must be lower or equal then 3.
It needs proving but I think its easy to prove. I don't have time now but someone else can use this to prove my solution.
I have tried to solve it a bit different.
Putting abc instead of 1 (because 1=abc) we get
$\displaystyle (a - abc + ac)(b - abc + ab)(c - abc + bc) $
$\displaystyle a(1 - bc + c)b(1 - ac + a)c(1 - ab + b) $
$\displaystyle (1 - bc + c)(1 - ac + a)(1 - ab + b) $
Now, lets consider next cases:
If a,b,c are all equal to 1 then $\displaystyle (1 - bc + c)(1 - ac + a)(1 - ab + b) \le 1
$ is true.
If one of them is equal to 1 and others not (for example let be $\displaystyle a = 1,bc = 1 $ )
we would get:
$\displaystyle (1 - bc + c)(1 - ac + a)(1 - ab + b) = $
$\displaystyle (1 - 1 + \frac{1}{b})(1 - \frac{1}{b} + 1)(1 - \frac{1}{c} + \frac{1}{c}) = $
$\displaystyle \frac{1}{b}(2 - \frac{1}{b}) $
Knowing that $\displaystyle c = \frac{1}{b} $ then $\displaystyle \frac{1}{b}(2 - \frac{1}{b}) $ is equal to:
$\displaystyle c(2 - c) =$
$\displaystyle 2c - c^2 $
Because c is positive, (2-c) can have only positive values 1 and 0 and for all other values (2-c) is always negative so
$\displaystyle 2c - c^2 \le 1$ is true.
Hope this solve the problem.
It's a pretty little problem, isn't it? It's relatively easy to use the Calculus to show that a=b=1 (where c=1/ab) is an absolute maximum for the expression (and the max = 1 so the expression is less than or equal to 1), but I doubt that this is the approach that the problem was designed for.
-Dan