Two questions about quadratic equations to be solved without using Cauchy-Schwarz inequality

$$Q1:$$ Let $$a_1,a_2,…,a_n$$ be non-zero real numbers and $$b_1,b_2,…,b_n$$ be real numbers. Find the discriminant of the quadratic equation $$(a_1x-b_1)^2 + (a_2x-b_2)^2 +…+(a_nx-b_n)^2 = 0$$ What can you say about the discriminant?
$$S1: \text{Discriminant},\Delta= 4\left[\left(\sum a_ib_i \right)^2 - \left(\sum a_i^2 \right) \left(\sum b_i^2 \right)\right]$$

$$Q2:$$ Let $$a,b,c$$ be real numbers. Consider the equation $$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a) = 0$$ Prove that the roots of the equation are always real. Further, show that the roots are equal $$\text{iff} \; a=b=c$$
$$S2: \Delta= 4\left((a^2+b^2+c^2)-(ab+bc+ca)\right)$$

One can easily say that for $$S1, \Delta \le 0$$ and for $$S2,\Delta \ge 0$$ using Cauchy-Schwarz Inequality. But the book (which I am following to study Algebra) haven’t told about the inequalities yet so we can’t use them, it’s the next topic.
How to do them without knowing the inequality?

• I have taken the liberty to change your title in order that it reflects the nature of the question. Do you agree ? Feb 17 at 13:18
• Yes, Thanks @JeanMarie. The title resonates more with the content now. Feb 17 at 14:36
• The crux behind this exercise is that $\sum_r x_r^2 \geq 0$ with equality when $x_r=0$. The first one is a proof of C-S that is also listed on Wikipedia. The second one is (weaker form of) another well-known and trivial inequality. Feb 19 at 3:25

Question Q1 :

A sum of squares is $$>0$$ unless all the squares are zero, in which case the sum is zero.

It is known that if a quadratic expression remains always $$>0$$, it cannot have roots, therefore its discriminant is $$<0$$.

What happens in the exceptional case where all the squares are zero ? (in which case the discriminant is $$0$$) ? It means that we have $$0=a_1x-b_1=a_2x-b_2=...$$ for the same value of $$x$$, otherwise said that we have the proportionnality $$x=\frac{b_1}{a_1}=\frac{b_2}{a_2}=...$$

(you haven't had to use Cauchy-Schwarz inequality ; on the contrary it is a way to establish it).

Question Q2 :

(Edit) Let us assume WLOG that $$a. Let

$$P(x):=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$$

• a) for any $$x$$ such that $$a, we have $$P(x)>0$$ (sum of products of $$>0$$ expressions) because all expressions $$(x-a)$$,$$(x-b)$$,$$(x-c)$$ are $$>0$$ .

• b) for any $$x$$ such that $$x, we have $$P(x)>0$$ as well (sum of products of $$<0$$ expressions) because all expressions $$(x-a)$$,$$(x-b)$$,$$(x-c)$$ are $$<0$$ .

• c) Besides $$P(b)=0+0+(b-c)(b-a)$$ which is $$<0$$ because $$b-c<0$$ and $$b-a>0$$.

Therefore, this quadratic $$P(x)$$ undergoes sign changes from $$>0$$ values to $$<0$$ values : consequently, it has necessarily real roots.

The case of equality $$a=b=c$$ is straighforward : cases a) and b) are similar and case c) becomes $$P(b)=0$$. I would day that in this case, the quadratic "hasn't enough space to change its sign, just the space to be $$0$$..."

A different, more complicated, solution involving calculus : consider the given expression $$P(x)$$ as the derivative of $$Q(x)=(x−a)(x−b)(x−c)$$ The signs of $$Q(x)$$ are $$-,+,-,+$$ on $$(-\infty,a),(a,b),(b,c),(c,+\infty)$$ resp. with necessarily one extrema in $$(a,b)$$ and one extrema in $$(b,c)$$ ; therefore variations of $$Q$$ are : ascending then descending then ascending, with local maxima at abscissa $$M$$ and local minima at abscissa $$m$$, with $$a≤M≤b≤m≤c$$ (the case of equality is evident from there), and $$M$$ and $$m$$ are, in a natural way, the roots of the derivative $$P(x)$$.

• For Q1: Hell yeah! that was trivial, I didn't notice that. It's a great way to prove C-S Inequality. For Q2: You cannot use Calculus either. Feb 17 at 14:46
• But I liked the calculus approach, It's amazing/cool/nice. Feb 17 at 14:54
• Well for $Q2$ I got something, $(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$ and when you expand the expression you get the required inequality and it is easy to see that the expression is $0$ iff $a=b=c$. Feb 17 at 15:39
• Sorry, I didn’t understand your edited answer Feb 18 at 2:44
• Thanks! I appreciate your efforts. Feb 19 at 2:46

For the last question, just draw the graph of the function $$f(x)=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}.$$

• Thanks, I already got the answer. See the last comment in Jean’s answer Feb 18 at 2:15