A question about triangles?

Question

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Answer 1

Use the law of cosines.

a^2 = b^2 + c^2 - 2bc cos A
b^2 = a^2 + c^2 - 2ac cos B
c^2 = a^2 + b^2 - 2ab cos C

Adding the equations

a^2+b^2+c^2 = 2(a^2+b^2+c^2)-2(bc cos A+ac cos B+ab cos C)
2(bc cos A+ac cos B+ab cos C) = a^2+b^2+c^2
a^2+b^2+c^2 = 2(bc cos A+ac cos B+ab cos C)

But cos <= 1. And angles A, B, and C are not all the same so the cosines can't all be 1.

Therefore

a^2 + b^2 + c^2 < 2(bc + ac + ab)

Answer 2

oh my math teacher had a poster that showed this.
basically if you place a square that is 3x3 and one that is 4x4 so that their corners make a 90 degree angle and then one that is 5x5 diagonay so it makes triangle with the other two. you get a 3,4,5 triangle. and counting the units of each one you'd get :
9 + 16 = 25
and then there was some other thing that wrapped everything up but I can't think what it is