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I just got a long expression: $$a^4 + b^4 + c^4 - 2b^2c^2 - 2a^2b^2 - 2a^2c^2$$ and I need to prove its less than zero for every $a$, $b$, and $c$ which are triangle sides

I really need tips how to handle such large expressions so they can be more useful for me. I know that I should probably get something sort of $(a+b+c)^2$ but I cant really find a way to do it , I tried some ways but a general rule that helps would be useful ...

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What are the inequalities that one can generate about the sides of a triangle? –  JB King Jun 8 at 22:27

3 Answers 3

up vote 10 down vote accepted

That expression can be factored into $-(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$. Now what can you say about each of the factors if $a,b,c$ are sidelengths of a triangle?

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That was fast factorization! :-) –  Sanath Devalapurkar Jun 8 at 22:24
how did you do that pls tip me i need to know how rather than resloving it –  user155994 Jun 8 at 22:25
@SanathDevalapurkar, it comes from Heron's formula –  Will Jagy Jun 8 at 22:25
@WillJagy Oh yes, it does! –  Sanath Devalapurkar Jun 8 at 22:26
thank you since a+b-c>0 everything is resloved –  user155994 Jun 8 at 22:29

Others have mentioned it came from Heron's formula. That expresses the area of a triangle as a function of the lengths $a$, $b$, and $c$ of the three sides. Notice that since "a straight line is the shortest distance between two points", we must have $a+b\ge c$, $b+c\ge a$, and $c+a\ge b$. If $a+b$ happens to be equal to $c$, so that the distance along one side plus the distance along the next side makes the trip in as little distance as going along the third side, then the three vertices of the triangle must be on a straight line, and the area is $0$. That means an expression for the area must have $a+b-c$ as a factor. For the same reason, it has $b+c-a$ and $c+a-b$ as factors. And since the area of a triangle is $0$ when all three sides have length $0$, it also has $a+b+c$ as a factor.

But $(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ is proportional to the fourth power of the lengths, whereas an area is proportional to the second power. Hence $\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$ is proportional to the area. Application to any particular triangle whose area and side lengths are known shows that the constant of proportionality must be $1/4$. So $$ \text{area of triangle} = \frac 1 4 \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$

That's Heron's formula. If you multiply out the polynomial under the radical, you get $$ -a^2-b^2-c^2+2ab+2bc+2ca. $$ Anyone who factors that polynomial instantly is probably someone who learned that by thinking about Heron's formula.

Heron's formula is also written in the form $$ \text{area}=\sqrt{s(s-a)(s-b)(s-c)} \text{ where }s=\frac{a+b+c}{2}. $$ The quantity $s$ is called the semiperimeter, obviously because it's half the perimeter.

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As others have mentioned,it follows in quite a straightforward way from Heron.However,we do what you initially suggested.This method,as you may already know,is called completing the square.My hint goes as follows:

Compare your expression with the expansion of $(a^2-b^2-c^2)^2$. What can you add or subtract from your expression to get a square?Finally,complete the factorization with repeated use of the identity $a^2-b^2=(a+b)(a-b)$

MOTIVATION: As you said,it was fairly easy to realize that we have an expression resembling $(a^2+b^2+c^2)^2$, but the minus signs made it clear that it wasn't what we should be looking for.The minus signs also motivated me to compare that expression with $(a^2-b^2-c^2)^2$.The rest was quite simple.

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