<s>[INST] If a,b and c are sides of a triangle, then prove that the following polynomial has no real roots This is the polynomial: $$a^2x^2+(b^2+a^2-c^2)x+b^2=0$$
Now this is my progress:
Assuming l,m, and n are sides of a triangle, then $$|m-n|\lt l\lt m+n$$
Also, if a second degree polynomial in the form $kx^2+px+q$ has real roots, then $$p^2-4kq\ge 0$$
 In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
 $$b^2+a^2-c^2\lt 2ab$$
 But by the first equation, $$a+b\gt c$$ 
$$a^2+b^2-c^2\gt 2ab$$ 
 Wich actually proves by contradiction there ARE real roots. What did I do wrong?
 [/INST] Nice exercise. It is well known (and not difficult to prove) that a second-degree polynomial has no real roots iff its discriminant is negative. The discriminant of the given polynomial is
$$ (b^2+a^2-c^2)^2-4a^2 b^2 = (a^2+2ab+b^2-c^2)(a^2-2ab+b^2-c^2) $$
or
$$ (a+b+c)(a+b-c)\color{red}{(a-b-c)}(a-b+c) $$
that is clearly negative by the triangular inequality or, equivalently, by Heron's formula.
</s>