<s>[INST] $4ac-b^2\leq 3a(a+b+c)$ in quadratic Suppose that the polynomial $(b+c)x^2+(a+c)x+(a+b)$ doesn't have real roots, where $a,b,c\in\mathbb{R}$. Prove that $4ac-b^2\leq 3a(a+b+c)$.
The quadratic not having real roots means that $$(a+c)^2-4(b+c)(a+b)<0$$ which translates to $(a^2+2ac+c^2)-4(b^2+ab+ac+bc)<0$, or 
$$a^2+c^2-2ac-4b^2-4ab-4bc<0$$
which is still quite far from the inequality in question. We need to eliminate the $c^2$ term, which might be possible using a square form like $(c-b)^2\geq 0$, but it doesn't really get us closer.
 [/INST] We have $$(a+c)^2<4(a+b)(b+c)$$ or
$$a^2+2ac+c^2<4b^2+4ab+4ac+4bc$$ or
$$c^2+a^2+4b^2-2ac-4bc+4ab<8ab+8b^2$$ or
$$(c-a-2b)^2<8b(a+b),$$ which gives
$$a^2(c-a-2b)^2\leq8a^2b(a+b).$$
In another hand, we need to prove that 
$$4ac-b^2\leq 3a^2+3ab+3ac$$ or $$a(c-a-2b)\leq2a^2+ab+b^2,$$ which is obvious for $a(c-a-2b)<0.$
Let $a(c-a-2b)\geq0.$
Thus, it's enough to prove that
$$a^2(c-a-2b)^2\leq(2a^2+ab+b^2)$$ for which it's enough to prove that
$$8a^2b(a+b)\leq(2a^2+ab+b^2)^2,$$ which is true by AM-GM:
$$(2a^2+ab+b^2)^2\geq4\cdot2a^2(ab+b^2)=8a^2b(a+b).$$
Done!
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