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

thank you in advance

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- Aug 31st 2007, 05:03 PMshoshotriple 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

thank you in advance - Aug 31st 2007, 11:50 PMtopsquark
It ain't elegant but it works:

$\displaystyle ab + ac + bc > 0$

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

1) $\displaystyle b + c > -\frac{bc}{a}$ if $\displaystyle a > 0$

or

2) $\displaystyle b + c < -\frac{bc}{a}$ if $\displaystyle a < 0$

or

3) $\displaystyle bc > 0$ if $\displaystyle a = 0$

Now:

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

So consider case 1):

1) $\displaystyle b + c > -\frac{bc}{a}$ if $\displaystyle a > 0$

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

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

If both b, c are negative then

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

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

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

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

Now if $\displaystyle a + b > 0$ then

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

is a contradiction because $\displaystyle c < 0$ and the RHS is negative.

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

Thus we are left with $\displaystyle a + b = 0$.

Thus

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

$\displaystyle c > 0$

another contradiction.

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 $\displaystyle a + b + c > 0$.

-Dan - Sep 1st 2007, 09:28 PMieatfood
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 AMtopsquark
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