Minimum value of $\dfrac{a+b+c}{b-a}$

Question

$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ . Find the minimum value of $\dfrac{a+b+c}{b-a}$

Attempt:

$b^2 \le 4ac$

$f(1) = a+b+c$

$f(0) = c$

$f(-1) = a-b+c$

$a>0$ and $c>0$

I am unable to utilize these things to find the minimum value of the expression $\equiv \dfrac{a+b+\frac {b^2}{4a}}{b-a}$

The answer given is $3$.

$c=0,\ b=\infty,\ a=-\infty$ where $\infty$ represents large numbers. When you sub in $\infty$, you get $\frac0{2\infty}$, which basically equals $0$. — Christopher Marley, May 08 '18 at 02:50
@ChristopherMarley but $a\ge 0$, otherwise $\lim_{x\to\infty}f(x)=-\infty$ — Masacroso, May 08 '18 at 02:53

score 9 · Answer 1 · answered May 08 '18 at 03:17

9

Alt. hint: $\;f(-2) = 4a - 2 b + c \ge 0 \iff 2a + c \ge 2(b-a)\,$, then:

$$ \dfrac{a+b+c}{b-a} = 1 + \frac{2a+c}{b-a} \;\ge\; 1 + 2 = 3 $$

answered May 08 '18 at 03:17

dxiv

76,497

What a fine trick! – Macrophage May 08 '18 at 03:19
Nice answer! Is there any motivation behind plugging in $x=-2$, or do you just try some small integer values of $x$ and see what crops up? – Carl Schildkraut May 08 '18 at 03:21
2

@Macrophage Thank you. This still leaves the OP with proving that the lower bound is in fact attained, but given the tone of the question I guess this won't be a problem. – dxiv May 08 '18 at 03:24
2

@CarlSchildkraut Thanks. It started with breaking a $,b-a,$ off the fraction, then trying to relate $,(2a+c),$ and $,\lambda (b-a),$ back to the quadratic somehow. I reckon this was a lucky shot, and/or this was meant to be a trick problem since the solution requires a lot less than the given assumptions. – dxiv May 08 '18 at 03:27

Macrophage · Accepted Answer · 2018-05-08T03:17:21.110

Actually, you are on the right track! Now let's complete it with only elementary calculus.

$$\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$$

Knowing that $b>a$, let's set $b=ka, k>1$. $$\frac{(2a+b)^2}{4a(b-a)}=\frac{(2+k)^2a^2}{4(k-1)a^2}=\frac{(2+k)^2}{4(k-1)}$$

Now you just need to find the minimum of this function for $k>1$. You can differentiate it and check that $k=4$ is a critical point, and you get $\frac{a+b+c}{b-a}=3$ there as a minimum.

Iceberry · Answer 3 · 2018-05-09T01:47:18.750

3

Rewrite the equation as $x^2+Bx+C$. Now, $B^2\le 4C$ So, the minimum value of $C=\dfrac{B^2}{4}$. Now, $\dfrac {a+b+c}{b−a} =\dfrac {1+B+(B^2/4)}{B−1}$. Find the minimum value of function.

edited May 09 '18 at 01:47

answered May 08 '18 at 17:27

Iceberry

634

Minimum value of $\dfrac{a+b+c}{b-a}$

3 Answers3