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

## 2 Answers

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]

?


============================================= • This is eye-opening. May I ask: how did you figure out the proper Hessian? – Lee David Chung Lin Oct 18 '18 at 23:20
• @LeeDavidChungLin I kept the first form as is and subtracted it off $t$ times the second form, requirement $t>0.$ The only value where such difference was positive semi-definite was $t=8$ – Will Jagy Oct 18 '18 at 23:39
• Thanks for the info. – Lee David Chung Lin Oct 18 '18 at 23:42

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!