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- Feb 14, 2012

- 3,632

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- Thread starter
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- #1

- Feb 14, 2012

- 3,632

- Admin
- #2

\(\displaystyle a>c\implies a^2>c^2\)

\(\displaystyle b>c\implies b^2>c^2\)

These two conditions also imply:

\(\displaystyle ab>c^2\)

Adding the three implications, we obtain:

\(\displaystyle a^2+ab+b^2>3c^2\)

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these last two, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

As we initially found, this is the result of assuming both \(a\) and \(b\) are greater than \(c\).

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- Feb 14, 2012

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Solution of other:

Adding $c$ to both sides, and square them yields $(a+c)^2\le 4c^2$.

By the triangle inequality we have $b<a+c$ and hence $b^2<(a+c)^2\le 4c^2$.

We then get $a^2+b^2<5c^2$, which reach to a contradiction and therefore, our assumption is wrong and $c$ is the length of the shortest side of the triangle.

- Mar 31, 2013

- 1,292

Hello Mark

\(\displaystyle a>c\implies a^2>c^2\)

\(\displaystyle b>c\implies b^2>c^2\)

These two conditions also imply:

\(\displaystyle ab>c^2\)

Adding the three implications, we obtain:

\(\displaystyle a^2+ab+b^2>3c^2\)

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these last two, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

As we initially found, this is the result of assuming both \(a\) and \(b\) are greater than \(c\).

You have assumed c being the shortest side and you have taken the given condition as well. So I am not convinced that

the proof is right. You have taken both the condition and assumption and proved it.

If I have missed something kindly let me know.

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- #5

I have assumed that \(c\) is the shortest side and shown how it leads to an implication provided both by the given and the triangle inequality. It seems to me this is sufficient. Is it not?Hello Mark

You have assumed c being the shortest side and you have taken the given condition as well. So I am not convinced that

the proof is right. You have taken both the condition and assumption and proved it.

If I have missed something kindly let me know.

- Mar 31, 2013

- 1,292

I have assumed that \(c\) is the shortest side and shown how it leads to an implication provided both by the given and the triangle inequality. It seems to me this is sufficient. Is it not?

But does not prove that this is not true if c is not the shortest side which we need to prove

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- #7

But does not prove that this is not true if c is not the shortest side which we need to prove

The triangle inequality implies:

\(\displaystyle a^2+2ab+b^2>c^2\)

And, we are given:

\(\displaystyle a^2+b^2>5c^2\)

Adding these, there results:

\(\displaystyle a^2+ab+b^2>3c^2\)

Now, suppose \(a=3b\) and \(c=2b\):

\(\displaystyle 9b^2+3b^2+b^2>12b^2\)

\(\displaystyle 13b^2>12b^2\)

This is true, even though \(c>b\). I'll try to come up with a sound solution.