Prove that for a triangle of side lengths a, b and c, a+b>c?

It's tempting to say it's obvious - but that's no proof!

The proof comes from Pythagorus' theorem - in a right angled triangle the square of the hypotenuse is equal to the sum of the squares on the other two sides.

Drop a line bisecting line c at a right angle from the angle when a and b join.

You now have two right angled triangles.

Call the height h and the bit of the line c that's in the 'b' right angled triangle length 'y'.

a^2 = h^2 + (c-y)^2

and

b^2 = h^2 + y^2

h is greater than zero, so we can say:

a^2 > (c-y)^2

b^2 > y^2

As we're talking about lengths we can ignore negative values and so say that menas:

a > c-y

b > y

Add the two equations up and that means:

a+b > c - y +y

a+b>c

if a+b = c then there is no triangle. We have (at best) a straight line with two segments a and b whose length end to end is c

if a+b < c then there is no triangle and the two segments a and b even when laid end to end can't even reach the length of c.

so, the only way to form a triangle is to have a + b > c