Question

Prove that in any triangle the length of a side is less than the half of the perimeter. In statement reason please.

Answer 1

Let
`a,b,c` be the sides of a triangle and
`p` its perimeter.

For any triangle, the below is true.

`a,b,c>0 \wedge a<b+c \wedge b<a+c \wedge <a+b`

`p=a+b+c`

So

`(p)/(2)=(a+b+c)/(2)`

Now, let's write the inequalities for the thesis and prove they're true.

`a<(a+b+c)/(2) \wedge b<(a+b+c)/(2) \wedge c<(a+b+c)/(2)\\\\2a<a+b+c \wedge 2b<a+b+c \wedge 2c<a+b+c\\\\a<b+c \wedge b<a+c \wedge c<a+b`

As a result, I got the exact same inequalities that are true for any triangle. Q.E.D.