Difficulf exercise! I can't do please help me

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February 16th 2006, 12:28 AM
luckyvan

Difficulf exercise! I can't do please help me

:eek: 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
February 16th 2006, 06:51 AM
DenMac21

Quote:

Originally Posted by luckyvan

:eek: 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

$(a - 1 + \frac{1}{b})(b - 1 + \frac{1}{c})(c - 1 + \frac{1}{a})$
$(\frac{{ab - b + 1}}{b})(\frac{{bc - c + 1}}{c})(\frac{{ac - a + 1}}{a})$
$\frac{{(ab - b + 1)(bc - c + 1)(ac - a + 1)}}{{abc}}$
$\frac{{(ab^2 c - abc + ab - b^2 c + bc - b + bc - c + 1)(ac - a + 1)}}{{abc}}$
$\frac{{(ab - b^2 c + 2bc - c)(ac - a + 1)}}{{abc}}$
$a + b + c - a^2 b - b^2 c - ac^2 + ab + bc + ac - 2$
$ab(1 - a) + bc(1 - b) + ac(1 - c) + a + b + c - 2$
$a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1) - 2$

Since a,b,c are all positive then
$a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1)$ 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.
February 16th 2006, 04:10 PM
DenMac21

I have tried to solve it a bit different.

Putting abc instead of 1 (because 1=abc) we get
$(a - abc + ac)(b - abc + ab)(c - abc + bc)$
$a(1 - bc + c)b(1 - ac + a)c(1 - ab + b)$
$(1 - bc + c)(1 - ac + a)(1 - ab + b)$

Now, lets consider next cases:
If a,b,c are all equal to 1 then $(1 - bc + c)(1 - ac + a)(1 - ab + b) \le 1<br />$ is true.

If one of them is equal to 1 and others not (for example let be $a = 1,bc = 1$ )
we would get:

$(1 - bc + c)(1 - ac + a)(1 - ab + b) =$
$(1 - 1 + \frac{1}{b})(1 - \frac{1}{b} + 1)(1 - \frac{1}{c} + \frac{1}{c}) =$
$\frac{1}{b}(2 - \frac{1}{b})$

Knowing that $c = \frac{1}{b}$ then $\frac{1}{b}(2 - \frac{1}{b})$ is equal to:
$c(2 - c) =$
$2c - c^2$

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
$2c - c^2 \le 1$ is true.

Hope this solve the problem.
February 16th 2006, 04:46 PM
topsquark

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
February 16th 2006, 05:50 PM
DenMac21

For what aproach is problem designed?

What is the solution?