What is the range of $λ$? Suppose $a, b, c$ are the sides of a triangle and no two of them are equal.
Let $λ ∈ IR$. If the roots of the equation $x^
2 + 2(a + b + c)x + 3λ(ab +
bc + ca) = 0$ are real, then what is the range of $λ$?
I got that $$λ ≤\frac{
(a + b + c)^
2}
{3(ab + bc + ca)}$$
After that what to do?
 A: There is a systematic way to get rid of the triangle condition by substituting 
$$a = x + y, b = y + z, c = z + x$$
which comes from the tangents to the incircle, explained e.g. here. Now the only condition on $x,y,z$ is that they are positive reals, and additionally they are distinct (because we want $a,b,c$ to be distinct).
The quadratic has real roots iff the discriminant is $\geq 0$, and as you say, if (writing in terms of the new variables)
$$\lambda \leq \frac{4(x+y+z)^2}{3(x^2+y^2+z^2) + 9(xy+yz+zx)}$$
Now we need to find the range of that expression for distinct real $x,y,z$.
Now let $a=x^2+y^2+z^2$, $b=xy+yz+zx$, and observe that $a> b$ from $(x-y)^2+(y-z)^2+(z-x)^2>0$, and moreover $a/b$ can take any value $\alpha> 1$. Hence, since our expression is
$$\frac{4}{3}\frac{a+2b}{a+3b} = \frac{4}{3}\frac{a/b+2}{a/b+3}$$
and we know the range of $a/b$, we get the range of $\lambda$ as well.
A: For a triangle with sides $a,b,c$ by triangle inequality, we have 
$$|a-b|<c$$ 
Squaring both sides we get,
$$(a-b)^2<c^2\tag{1}$$
Similarly, 
$$(b-c)^2<a^2\tag{2}$$
And
$$(c-a)^2<b^2\tag{3}$$ 
Adding $(1),(2)$ and $(3)$, we get
$$a^2+b^2+c^2 <2(ab+bc+ca) \Longleftrightarrow (a+b+c)^2 <4(ab+bc+ca)\tag{4}$$
From $(4)$, we get $\lambda <\dfrac{4}{3}$

To see why the upper bound is the supremum consider a degenerate triangle with sides $(a,a,c)$,(where $c \rightarrow 0$) and permutations.
For example, when $(a,b,c)=(1,1, 0.01)$, then $\lambda \leq \dfrac{4.0401}{3.0600} \approx 1.321 $
