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$Q1:$ 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?

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    $\begingroup$ I have taken the liberty to change your title in order that it reflects the nature of the question. Do you agree ? $\endgroup$
    – Jean Marie
    Feb 17 at 13:18
  • $\begingroup$ Yes, Thanks @JeanMarie. The title resonates more with the content now. $\endgroup$ Feb 17 at 14:36
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    $\begingroup$ 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. $\endgroup$
    – Sahaj
    Feb 19 at 3:25

2 Answers 2

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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 $a<b<c$. Let

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

  • a) for any $x$ such that $a<b<c<x$, 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 $x<a<b<c$, 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 $P(x)$.

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    $\begingroup$ 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. $\endgroup$ Feb 17 at 14:46
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    $\begingroup$ But I liked the calculus approach, It's amazing/cool/nice. $\endgroup$ Feb 17 at 14:54
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    $\begingroup$ 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 $a=b=c$. $\endgroup$ Feb 17 at 15:39
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    $\begingroup$ Sorry, I didn’t understand your edited answer $\endgroup$ Feb 18 at 2:44
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    $\begingroup$ Thanks! I appreciate your efforts. $\endgroup$ Feb 19 at 2:46
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For the last question, just draw the graph of the function $$f(x)=\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}.$$

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

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