# Quadratic equation with tricky conditions. Need to prove resulting inequalities.

The roots of the quadratic equation $ax^ 2-bx+c=0,$ $a>0$, both lie within the interval $[2,\frac{12}{5}]$. Prove that: (a) $a \leq b \leq c <a+b$. (b) $\frac{a}{a+c}+\frac{b}{b+a}>\frac{c}{c+b}$

So we can use the quadractic formula and obtain $2 \leq \frac{b \pm \sqrt{b^2-4ac}}{2a} \leq \frac{12}{5}$

and as $a>0$, this implies that

$4a \leq b \pm \sqrt{b^2-4ac} \leq \frac{24a}{5}$.

Then we can divide this into two cases (plus or minus) --- But how can we manipulate it to yield the required result?

Another idea is to make use of the fact that $b^2-4ac$ is non-negative

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As shown in other answers, let $\alpha,\beta$ be the two real roots of $ax^2-bx+c=0$, i.e. $$ax^2-bx+c=a(x-\alpha)(x-\beta)\iff \alpha+\beta=\frac{b}{a},\ \alpha\beta=\frac{c}{a}.$$

Since $\alpha,\beta\in[2,2.4]$ and $a>0$, $$\frac{b}{a}=\alpha+\beta\ge 4\Rightarrow a\le \frac{b}{4}<b;$$ $$\frac{b}{c}=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{\alpha}+\frac{1}{\beta}\le 1\Rightarrow b\le c;$$ and $$(\alpha-1)(\beta-1)<2\Rightarrow 1+\frac{b}{a}=\alpha+\beta+1>\alpha\beta=\frac{c}{a}\Rightarrow c< a+b.$$ Therefore, $$\frac{a}{a+c}+\frac{b}{a+b}\ge \frac{a}{a+c}+\frac{b}{a+c}=\frac{a+b}{a+c}>\frac{c}{a+c},$$ where the first inequality is due to $a,b>0$ and $0\le b\le c$ and the second one is due to $a+b>c$ and $a+c>0$.

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Hints: let $\,\alpha\,,\,\beta\,$ be the quadratic's roots, then Viete's formulas give

$$\alpha+\beta=\frac{b}{a}\;,\;\;\alpha\beta=\frac{c}{a}$$

Suppose $\,2\le \alpha\le\beta\le 2.4\,$ , then for example

$$a=\frac{b}{\alpha+\beta}\le b\ldots$$

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Note that the equation is $ax^2 \boxed{-} bx+c=0$. So you mean $\alpha +\beta=\frac{-b}{a}$ – John Marty Apr 10 '13 at 2:33
But even so how do you know that $\frac{-b}{\alpha+\beta} \leq b$?; ($b$ isn't necessarily positive) – John Marty Apr 10 '13 at 2:37
I see, $b$ is positive, thanks. I'm getting much closer to solving this – John Marty Apr 10 '13 at 2:44
$$\alpha\,,\,\beta\in [2,2.4]\implies \alpha+\beta>1\implies \frac{b}{\alpha+\beta}<b$$ – DonAntonio Apr 10 '13 at 10:40
Dont mention it, @JohnMarty , and there are two main kinds of idiocies trhat are widespread in this site: not to learn from one own mistakes and/or not trying to do some self effort after being helped several times in the same subject. I'm sure both do not apply to you. – DonAntonio Apr 10 '13 at 11:18