Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$.

Question

$a$, $b$, $c$ are three positives such that $ab < b^2 < 4ca$. Prove that $$\large \dfrac{a + b + c}{b - a} > 3.$$

I can't think of a way to get around this problem. Although I can see that based on the given condition, $ax^2 + bx + c = 0$ has no roots, which adds almost no information whatsoever.

If you have written an answer below, thanks for that!

In the title, you have $b-a$ as the denominator, but it's $c-a$ in the body of your question. I'm guessing one of those is a typo. — Robert Howard, Mar 18 '19 at 04:04
So, this is equivalent to showing $\frac{2a+c}{b-a}>2\to2a+c>2b-2a\to\frac{4a+c}2>b$. But, by AM-GM, we know that $\frac{4a+c}2\geq\sqrt{4ac}>b$, so we are done. — Rushabh Mehta, Mar 18 '19 at 04:07
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score 4 · Accepted Answer · answered Mar 18 '19 at 05:12

As per my comments, here's how I would approach this question. Note that since $a,b,c>0$, $b>a\to b-a>0$. So, we know via the AM-GM inequality that $$\frac{4a+c}2\geq\sqrt{4ac}>b$$$$4a+c>2b\to 2a+c>2b-2a$$$$\frac{2a+c}{b-a}>2$$$$1+\frac{2a+c}{b-a}>3$$$$\frac{a+b+c}{b-a}>3$$

jacky · Answer 2 · 2019-03-18T05:57:15.997

2

Assuming $f(x) = ax^2+bx+c>0$ for all real $x$ and

$a>0$ and $(b-a)>0$

put $x=-2,$ We get

$f(-2) =4a-2b+c> 0\Rightarrow 2a+c>2(b-a)$

$\Rightarrow \displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}>1+2=3.$

edited Mar 18 '19 at 05:57

answered Mar 18 '19 at 05:25

jacky

5,194

1

+1, shortcut: $4a-2b+c>0 \iff a+b+c>3(b-a)$ – farruhota Mar 18 '19 at 08:19

score 1 · Answer 3 · answered Mar 19 '19 at 11:58

1

$\displaystyle a+b+c>a+b+\frac{b^2}{4a}=\bigg(4a+\frac{b^2}{4a}\bigg)+b-3a\geq 3(b-a)$

$$\displaystyle \frac{a+b+c}{b-a}>3$$

answered Mar 19 '19 at 11:58

jacky

5,194

Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$.

3 Answers3