Thread: BMO 2010/11 question on inequalities

Originally Posted by Ackbeet

However, it's still not clear to me how you get that all three have to be less than one. Let's say you assume a < b < c. Then you can easily show that a and b must be less than one. Why does c have to be less than one?

The key condition a + b > c tells you that if c > 1 then a + b >1. But then — contradiction.

Originally Posted by Opalg

The key condition a + b > c tells you that if c > 1 then a + b >1. But then — contradiction.

Nice!

Originally Posted by Ackbeet

However, it's still not clear to me how you get that all three have to be less than one. Let's say you assume a < b < c. Then you can easily show that a and b must be less than one. Why does c have to be less than one? And if you assume a different ordering, well, that's a separate case. In each possible ordering, you still get one side that could, theoretically, be greater than one. The fact that they're disjoint cases is neither here nor there.

Incidentally, I think it's more realistic to allow equalities in your assumption. That is, you really should assume

Thanks for the comment.

*First (from last), the reason that I assumed without possible equality is that the side lengths of the triangle are not equal (unless it's degnerate in which case the result is still true).

*There's no need to consider all orderings (six of them), if I wrote instead the argument would be the same. That's what I mean by "without loss of generality".

Originally Posted by Ackbeet

However, it's still not clear to me how you get that all three have to be less than one. Let's say you assume a < b < c. Then you can easily show that a and b must be less than one. Why does c have to be less than one? And if you assume a different ordering, well, that's a separate case. In each possible ordering, you still get one side that could, theoretically, be greater than one. The fact that they're disjoint cases is neither here nor there.

Incidentally, I think it's more realistic to allow equalities in your assumption. That is, you really should assume

we have and . I agree that the cases where equalities occur should also be considered but that can be done separately taking and etc.

EDIT: didn't see that others had already answered the question.

Originally Posted by melese

Thanks for the comment.

*First (from last), the reason that I assumed without possible equality is that the side lengths of the triangle are not equal (unless it's degnerate in which case the result is still true).

*There's no need to consider all orderings (six of them), if I wrote instead the argument would be the same. That's what I mean by "without loss of generality".

Arguing in a "without loss of generality" manner implies that you're setting up a by-cases style argument. That is, in Case 1, you get result A; in Case 2, you get result A, and in all the remaining cases, you get result A. The "without loss of generality" part means that it is immediately clear by symmetries in the argument that all the cases work out the same. However, a < b < c, in your argument (leaving out Opalg's clincher here), means that a<1 and b<1. It does not imply that c<1. The other cases do not similarly show, via your argument, that a<1 and b<1. Hence, you cannot use the argument by cases in this fashion.

Now, if you employ Opalg's additional line of reasoning in Post # 16, you get that all three are less than 1 in just one case, and the three variables appear symmetrically in the proof. Then you can use the without loss of generality to save yourself the trouble of writing the other cases.

Concerning equality: I don't see why isosceles triangles should be excluded from the proof. That's not a degenerate case. But not allowing equality in your inequalities leaves out the isosceles case, and for that matter the equilateral case. You could do those separately, I suppose. But Opalg's reasoning gets you there using the more general (in)equalities.

Cheers.

Originally Posted by Also sprach Zarathustra

A little progress...

ab+bc+ac=1

1=b(a+c)+ac>b^2+ac

b^2+ac<1

0<ac<1-b^2=(1-b)(1+b)

b\in(0,1)

with the same argument like above one(?) can conclude that a,b,c\in(0,1)

Since a lot has been written and said about this inequality and since Zarathustra just missed a crucial multiplication to give the final solution ,i think i must give this multiplication ,although a long time has elapsed since this problem appeared in this subforum

The multiplication is :

(1-a)(1-b)(1-c)>0 <=> abc +a+b+c <2 <=> (a+1)(b+1)(c+1)<4