4

Suppose $a,b,c$ are real numbers such that $a+b+c>0$, $ab+bc+ca>0$, and $abc>0$. Prove by contradiction that $a,b,c>0$.

I have tried to solving it case by case like:
case $1$: $a,b,c<0$;
case $2$: exactly $1$ or $3$ variable(s) from $a,b,c$ is $<0$, the remainings are $>0$;
case $3$: exactly $2$ variables from $a,b,c$ are $<0$, the remaining is $>0$.

rrozamelo
  • 41

3 Answers3

3

So $a, b, c$ are roots of the cubic $$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$ by Descartes' rule of signs, it has no negative roots and $0$ isn't a root. Hence all the roots must be positive, as they are given to be real...

If you want a proof strictly by contradiction, suppose there is a negative root, then all terms in the polynomial are negative, so it cannot sum to zero...

Macavity
  • 46,381
1

It is possible to show it at once, that is, avoiding cases.

Assume that $a\le b\le c$ and $a\le 0$. Since $abc>0$ then $a<0$ and $bc<0$. Now, $ab+ac+bc=a(b+c)+bc<a(b+c)<a(a+b+c)<0$

ajotatxe
  • 65,084
1

Assume $c<0$

Multiply inequality (2) by c,

$c(ab+bc+ca)<0$

$abc+(a+b)c^2<0$

Since $abc>0$ is given and $c^2>0$ as $c<0$, from the above inequality,

$a+b<0$

$a+b+c<c<0$

But $a+b+c>0$ (given)

This is contradiction. Hence, $c\ge 0$.

Similarly, we can derive $a\ge0$ and $b\ge0$

Deepak Gupta
  • 570