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?
 A: 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.
A: The following answers the What did I do wrong? part of the question

In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
   $$b^2+a^2-c^2\lt 2ab$$

The two inequalities are not equivalent, since $b^2+a^2-c^2$ can be either positive or negative (remember that it is $0$ for right triangles by Pythagora's). What the first inequality is equivalent with, instead, is:
$$\left|b^2+a^2-c^2\right|\lt 2ab \quad \iff \quad -2ab \lt b^2+a^2-c^2\lt 2ab \tag{1}$$

But by the first equation, $$a+b\gt c$$ 
   $$a^2+b^2-c^2\gt 2ab$$ 

The second inequality does not follow, in particular $2ab$ has the wrong sign. Instead, if you square the first inequality you get: $$a^2+b^2 - c^2 \gt -2ab \tag{2}$$
If you square $|a-b| \lt c$ though, then you get:
$$a^2+b^2 - c^2 \lt 2ab \tag{3}$$
Note that the last two inequalities $(2),(3)$ put together give $(1)$ and therefore complete the proof.
