Need algebra tip about $a^4 + b^4 + c^4 - 2b^2c^2 - 2a^2b^2 - 2a^2c^2$ for sides of a triangle 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 ...

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.
