For all triangles,

by cosine rule,

c²=a²+b²-2abcosC

(a+b)²=a²+2ab+b²

a+b=√(a²+2ab+b²)

c=√(a²+b²-2abcosC)

now, a+b is definitely larger than c, as the maximum possible value of -2abcosC is less than 2ab, as angle C is smaller than 180 but larger than 0.

]]>by pythagoras theorem,

a²+b²=c²

c=√(a²+b²) ----(1)

(a+b)²=a²+2ab+b²

a+b=√(a²+2ab+b²) ----(2)

comparing (1) and (2),

one can see that

a+b>c

Or not ...

]]>The Hero's formula for area of a triangle is

Area = √(s(s-a)(s-b)(s-c)

where a, b, c are the sides and s = (a+b+c)/2

Therefore, Area = √[(11/2)(11/2 - 2)(11/2 - 3) (11/2 - 6)]

11/2 - 6 is a negative number,

So the Area of the triangle is an imaginary number ]]>

Just interested.]]>

AB < (4+8)/2

AB < 12/2

AB < 6]]>

AB + BC > AC

AC + BC > AB

AB + AC > BC

You have given BC=4 and AC = 8 - AB

Therefore, the required inequality is

AB < BC + AC

AB < BC + 8 - AB

2 AB < BC + 8

AB < (BC + 8)/2

Help?

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