# Concluding three statements regarding $3$ real numbers.

$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$

Conclusion $I.)\ 1<b<3$

Conclusion $II.)\ 2<a<3$

Conclusion $III.)\ 0<c<1$

Options

By the given statements

$\color{green}{a.)\ \text{Only conclusion$I$can be derived}}$.

$b.)\$ Only conclusion $II$ can be derived.

$c.)\$ Only conclusion $III$ can be derived.

$d.)\$ Conclusions $I,\ II,\ III$ can be derived.

$e.)\$ None of the three conclusions can be derived.

$\quad\\~\\$

I tried $(a+b+c)^2=36 \implies a^2+b^2+c^2=18$

and found that $(a,b,c)\rightarrow \{(-1,1,4),(-3,0,3)\}$ satisfies the two conditions

$a^2+b^2+c^2,\ a<b<c$ but not this one $a+b+c=6$

I thought a lot but can't find any suitable pairs.

I look for a simple and short way.

I have studied maths upto $12$th grade.

• But if to solve a system of Diophantine equations? \left\{\begin{aligned}&a+b+c=6x\\&ab+ac+bc=9x^2\end{aligned}\right. But there is one number always turns negative. – individ Jun 29 '15 at 11:40
• I'm in the sign was wrong. There is a solution. This. $$a=3p^2$$ $$b=3s^2$$ $$c=3(p+s)^2$$ $$x=p^2+ps+s^2$$ Or so. $$a=4s^2$$ $$b=9p^2+6ps+s^2$$ $$c=9p^2-6ps+s^2$$ $$x=3p^2+s^2$$ – individ Jun 29 '15 at 14:54

I shall show that only Conclusion I is correct. For a fixed $b$, we have

$a+c=6-b$ and $ac=9-(a+c)b=9-(6-b)b=(b-3)^2$. Hence, the quadratic polynomial

$x^2-(6-b)x+(b-3)^2$ has two distinct real roots $x=a$ and $x=c$. Therefore,

the discriminant $(6-b)^2-4(b-3)^2=3b(4-b)$ of this quadratic is strictly

positive (hence, $0<b<4$). Furthermore, the roots of the quadratic are

$a=\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}$ and $c=\dfrac{(6-b)+\sqrt{3b(4-b)}}{2}$. As

$a<b<c$, we must have $\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}<b<\dfrac{(6-b)+\sqrt{3b(4-b)}}{2}$, or equivalently,

$-\sqrt{3b(4-b)}<3b-6<+\sqrt{3b(4-b)}$. Hence,

$\sqrt{3}|b-2|<\sqrt{b(4-b)}$, or $3(b-2)^2<b(4-b)$. Ergo, $4(b-1)(b-3)<0$.

That means $1<b<3$.

Note that $b=2$ gives a solution $(a,b,c)=(2-\sqrt{3},2,2+\sqrt{3})$.

Consequently, Conclusions II and III are false. In fact, we can prove that

$0<a<1<b<3<c<4$. It might be a good exercise for you to show that $0<abc<4$.

$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$

Conclusion I.) $1<b<3$

Conclusion II.) $2<a<3$

Conclusion III.) $0<c<1$

Let $\{p,q,r\}\in \mathbb{R}: a=\min(p,q,r),\ c=\max(p,q,r), b=\max(\min(p,q),\min(q,r),\min(r,p))$. Let's solve a system \begin{align} p+q+r&=6 ,\\ pq+qr+rp&=9 \end{align} for $q$ and $r$ in terms of $p$: \begin{align} q(p)&=3-\frac{1}{2} p-\frac{1}{2} \sqrt{3p (4-p)}, \\ r(p)&=3-\frac{1}{2} p+\frac{1}{2} \sqrt{3p (4-p)}. \end{align} The solution suggests that $0\le p\le 4$ (to ensure real values).

This diagram illustrates relationship between $p,q$ and $r$:

Analysis of $q(p)$ and $r(p)$ (boundary end extreme points) shows that \begin{align} a&=\begin{cases} p,& 0<p<1\\ q(p),& 1<p<4 \end{cases} \\ b&=\begin{cases} q(p),& 0<p<1\\ p,& 1<p<3\\ r(p),& 3<p<4 \end{cases} \\ c&=\begin{cases} r(p),& 0<p<3\\ p,& 3<p<4\\ \end{cases} \end{align}

And

$0<a<1$

$1<b<3$

$3<c<4$: