SOLUTION: Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c). What about a^2, b^2, and c^2?

Let assume that  

     +  <= .    (1)


where "a", "b" and "c" are the sides of a triangle.


I want to lead it to CONTRADICTION.


Indeed, square both sides of (1). You will get


    a +  + b <= c,   or


    a + b <= c - .


Then even more so

    a + b < c.


But it contradicts to the triangle inequality  a + b > c.


Thus we proved that  a + b > c   IMPLIES    +  > .


It works for any combinations of sides of a triangle.


Thus we proved that the values  ,  ,   satisfy all triangle inequalities,

if  "a",  "b"  and  "c"  are the sides of a triangle.


It implies that if  "a",  "b"  and  "c"  are the sides of a triangle, then  ,  ,  and    form a triangle, too.