$4ac-b^2\leq 3a(a+b+c)$ 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.

in standard order $a^2, b^2, c^2, bc, ca, ab,$ you are given
$$ -a^2 + 4b^2 - c^2 + 4bc +2ca+ 4 ab > 0 $$
The following is positive semidefinite (rank one Hessian matrix)
$$ 25a^2 + 4b^2 + c^2 - 4bc -10ca+ 20 ab \geq 0 $$
It is, in fact, simply $(5a+2b-c)^2$
Add to get
$$ 24a^2 + 8b^2 + 0 c^2 + 0bc -8ca+ 24 ab > 0 $$
or
$$ 24a^2 + 8b^2 -8ca+ 24 ab > 0 $$
Divide by $8$ to get
$$ 3a^2 + b^2 -ca+ 3 ab > 0 $$
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As to how to find this: the (a,b,c) Hessian matrix of $$ t(3a^2 + b^2 -ca+ 3 ab) - (-a^2 + 4b^2 - c^2 + 4bc +2ca+ 4 ab), $$
restricted to $t>0,$ turned out to have determinant $$ -2(t+1)(t-8)^2 $$
so the only positive $t$ giving a non-negative determinant was $t=8$
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PARI/GP is free software, 

? a = [ 6,3,-1; 3,2,0; -1,0,0]
%1 = 
[ 6 3 -1]

[ 3 2  0]

[-1 0  0]

? b = [ -2,4,2;4,8,4;2,4,-2]
%2 = 
[-2 4  2]

[ 4 8  4]

[ 2 4 -2]


? h = t * a - b
%8 = 
[6*t + 2 3*t - 4 -t - 2]

[3*t - 4 2*t - 8     -4]

[ -t - 2      -4      2]


? d = matdet(h)
%11 = -2*t^3 + 30*t^2 - 96*t - 128
? factor(d)
%12 = 
[t - 8 2]

[t + 1 1]

? 

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