Here is a problem which was given in a Norwegian mathematical contest:

 

Suppose [MATH]a[/MATH], [MATH]b[/MATH] and [MATH]c[/MATH] are real numbers such that [MATH]ab + bc + ca > a + b + c > 0[/MATH]. Show that [MATH]a + b + c > 3[/MATH].

Since nobody have replied yet, I will give hints to two fundamentally different solutions to the problem:

 

(1) Let x = b + c, y = bc, find limitations for x and y, and then find a lower limit for a + x. (This is my own solution to the problem.)

 

(2) Just try to figure something out relying on facts such that (a - b)^2 + (b - c)^2 + (a - c)^2 >= 0 and (a + b + c)^2 = ... (The official solution.)

 

It is possible that a solution based on for instance Vietas formulas would work too, but I have not found any.

 

In the contest the whole lot of three people managed to find a proof, so it should not be all that difficult.