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

$$a$$, $$b$$, $$c$$ 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 at 4:04
• 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. – Don Thousand Mar 18 at 4:07
• @DonThousand : Why not writing it as an answer? – trancelocation Mar 18 at 5:05
• Eh, I don't really like answering qs anymore. I tend to get obsessive over rep-hunting, so I tend to avoid it. – Don Thousand Mar 18 at 5:07
• @DonThousand : It is not rep-hunting. People who use the board to look for answers should find the answers in the answers section and should not be 'forced' to browse comments. That's why I recommend writing answers instead of comments even if they are very short or very simple. – trancelocation Mar 18 at 5:12

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 $$\frac{4a+c}2\geq\sqrt{4ac}>b$$$$4a+c>2b\to 2a+c>2b-2a$$$$\frac{2a+c}{b-a}>2$$$$1+\frac{2a+c}{b-a}>3$$$$\frac{a+b+c}{b-a}>3$$

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

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

put $$x=-2,$$ We get

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

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

• +1, shortcut: $4a-2b+c>0 \iff a+b+c>3(b-a)$ – farruhota Mar 18 at 8:19

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

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