Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 1 down vote accepted

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 $a+c>0$.

share|cite|improve this answer

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


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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.