# triple inequalities

• Aug 31st 2007, 05:03 PM
shosho
triple inequalities
i really dontknow this questoin any help would be great

prove that if real numbers a,b,c satisfy

a + b + c>0, ab +ac + bc>0 , abc>0

then each of a,b,c is positive

• Aug 31st 2007, 11:50 PM
topsquark
Quote:

Originally Posted by shosho
i really dontknow this questoin any help would be great

prove that if real numbers a,b,c satisfy

a + b + c>0, ab +ac + bc>0 , abc>0

then each of a,b,c is positive

It ain't elegant but it works:

$ab + ac + bc > 0$

$a(b + c) + bc > 0$

1) $b + c > -\frac{bc}{a}$ if $a > 0$
or
2) $b + c < -\frac{bc}{a}$ if $a < 0$
or
3) $bc > 0$ if $a = 0$

Now:
3) is the easiest case. $abc > 0$ and $a = 0$ are contradictory. Thus $a \neq 0$.

So consider case 1):
1) $b + c > -\frac{bc}{a}$ if $a > 0$

We know that $abc > 0$ and a is positive, so $bc > 0$. This means that either both b, c are positive or both b, c are negative.

If both b, c are positive then $b + c > -\frac{bc}{a}$ is trivial since the RHS is negative, and we have no contradiction.

If both b, c are negative then
$b + c > -\frac{bc}{a}$

$\frac{bc}{a} + c > -b$

$c \cdot \left ( \frac{b}{a} + 1 \right ) > -b$

$c \cdot \left ( \frac{a + b}{a} \right ) > -b$

Now if $a + b > 0$ then
$c > \left ( \frac{a}{a + b} \right ) \cdot -b$
is a contradiction because $c < 0$ and the RHS is negative.

If $a + b < 0$ and we know that c is negative then $a + b + c > 0$ cannot be true because $(a + b) + c < 0$, a contradiction.

Thus we are left with $a + b = 0$.
Thus
$c \cdot \left ( \frac{a + b}{a} \right ) > -b$

$c > 0$

Thus both b and c cannot be negative.

Thus case 1) says that if a is positive, so are both b and c.

Case 2) works in a similar fashion to show that, at best, all three of a, b, and c must be negative. (I leave this for you to show.) This again contradicts $a + b + c > 0$.

-Dan
• Sep 1st 2007, 09:28 PM
ieatfood
hey shosho seems to understand this problem.. but im not quite sure i do.
i dont understand the contradiction bit.. could you explain more? it seems like a reeli hard question
thanks
• Sep 2nd 2007, 05:17 AM
topsquark
Quote:

Originally Posted by ieatfood
hey shosho seems to understand this problem.. but im not quite sure i do.
i dont understand the contradiction bit.. could you explain more? it seems like a reeli hard question
thanks

The basic idea behind what I was doing is this:
Given a real number a, we know that only one of the following possibilities is true:
a < 0
a = 0
a > 0

I used this idea in two places in the proof of case 1): First in showing that if a is positive then b,c must both be either positive or negative. I showed that there is no contradiction if both b, c are positive. Then I showed that if b,c are both negative that a contradiction exists by looking at a + b in reference to >,=,<. There was a contradiction in all three cases, so I showed that a + b, a real number, is neither >, =, or < 0, an impossibility.

-Dan