<s>[INST] If $a$, $b$, $c$ are distinct sides of a triangle and the roots of $x^2+2(a+b+c)x+3m(ab+bc+ca)=0$ are real, show that $m<4/3$ if $m$ is real 
Let $a$, $b$, $c$ be the sides of a triangle such that all are distinct. If the roots of the equation $$x^2+ 2(a+b+c)x +3m (ab+bc+ac) =0$$ are real, then show that $m<\frac43$ if $m $ is a real number.

I've been trying to solve this question since a long time but I'm unable to do so. I don't understand how to use the "sides of the triangle condition" except that they show $a$, $b$, $c$ are positive real numbers. 
Would someone please help me to prove the required result?
 [/INST] The roots of the equation$$x^2+ 2(a+b+c)x +3m (ab+bc+ac) =0$$are real, so we have
$$4(a+b+c)^2-4\cdot 3m(ab+bc+ca)\ge 0,$$
i.e.
$$m\le \frac{(a+b+c)^2}{3(ab+bc+ca)}\tag1$$
Let $$x=a+b-c\gt 0,\ \ y=b+c-a\gt 0,\ \ z=c+a-b\gt 0$$
Then, $(1)$ can be written as
$$m\le \frac 43\cdot \frac{(x+y+z)^2}{x^2+y^2+z^2+3xy+3yz+3zx}\tag2$$
Now, we have
$$x^2+y^2+z^2+3xy+3yz+3zx-(x+y+z)^2=xy+yz+zx\gt 0$$
from which we have
$$\frac{(x+y+z)^2}{x^2+y^2+z^2+3xy+3yz+3zx}\lt 1\tag3$$
It follows from $(2)(3)$ that
$$m\lt\frac 43$$
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