- #1
Danijel
- 43
- 1
The problem: prove that if 1/a + 1/b + 1/c=1 and a,b,c are positive numbers then (a-1)(b-1)(c-1)>=8
I've tried it myself but couldn't do it. I have tried going backwards i.e. :
(a-1)(b-1)(c-1)>=8
(fast forward) abc -(ab+bc+ac) +a+b+c -1 >=8
Now from 1/a + 1/b + 1/c = 1 we have that ab+bc+ac/abc=1 thus :
ab + bc + ac = abc , so abc -(ab+bc+ac) +a+b+c -1>=8 is really a+b+c -1>=8
that is a+b+c >=9 . So If I could somehow prove that a+b+c>9 then I could work backwards and prove the former statement.
But the only progress I've made is :all of a,b,c must be greater than 1 because if they weren't then any of the 1/a,1/b,1/c would be greater then 1 thus we would need negatives(and a,b,c are positive) to "bring back" the value to 1, nor can they equal to 1 because the same thing would happen(one of 1/a,1/b,1/c would be equal to 1 and we would need negatives)so a>1, b>1, c>1 adding these three together we get a+b+c >3 ,and I need a+b+c+>=9.
Sorry for the long post. Could anyone give me some directions?
Thanks,
Daniel
I've tried it myself but couldn't do it. I have tried going backwards i.e. :
(a-1)(b-1)(c-1)>=8
(fast forward) abc -(ab+bc+ac) +a+b+c -1 >=8
Now from 1/a + 1/b + 1/c = 1 we have that ab+bc+ac/abc=1 thus :
ab + bc + ac = abc , so abc -(ab+bc+ac) +a+b+c -1>=8 is really a+b+c -1>=8
that is a+b+c >=9 . So If I could somehow prove that a+b+c>9 then I could work backwards and prove the former statement.
But the only progress I've made is :all of a,b,c must be greater than 1 because if they weren't then any of the 1/a,1/b,1/c would be greater then 1 thus we would need negatives(and a,b,c are positive) to "bring back" the value to 1, nor can they equal to 1 because the same thing would happen(one of 1/a,1/b,1/c would be equal to 1 and we would need negatives)so a>1, b>1, c>1 adding these three together we get a+b+c >3 ,and I need a+b+c+>=9.
Sorry for the long post. Could anyone give me some directions?
Thanks,
Daniel