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If $\frac{x^2+ax+3}{x^2+x+a}$ takes all real values for possible real values of $x$, then prove that $4a^3+39<0$. Here is how I approached it.

Let $$\frac{x^2+ax+3}{x^2+x+a}=y$$

Then, $$(y-1)x^2+(y-a)x+(ay-3)=0$$ We want all those $y$, for which there is a real $x$, that is, we want $y$ such that this quadratic has real roots. So, the discriminant $\Delta \geq 0$. $$(y-a)^2-4(y-1)(ay-3) \geq 0$$

On simplifying, we obtain $$(1-4a)y^2+(2a+12)y+(a^2-12) \geq 0$$ We want to find those $a$ for which this is true for all $y$. So, the discriminant $\Delta \leq 0$ (so that the parabola never crosses the $x$ axis.) and $(1-4a)>0$ (so that it faces upwards and is always above the $x$ axis.)

This gives $$(2a+12)^2-4(1-4a)(a^2-12) \leq 0$$ $$(a+6)^2-(1-4a)(a^2-12) \leq 0$$ which simplifies to $$a^3-9a+12 \leq 0$$ which is not what I set out to achieve. Where did I go wrong?

And is there any other method to do this? (Perhaps Calculus based?)

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    $\begingroup$ The only logical error I can see in your argument, which probably doesn't end up making a difference in the result, is that you need to account separately for what happens when the quadratic term is zero, as the quadratic formula is not valid in that case. Your result is not necessarily wrong, as it implies 4a^3 + 39 < 0. The graphs of these two functions show this. An alternative method would be to say that since the function tends to 1 at $\pm \infty$, the condition will be satisfied if and only if: (1) the function tends to $\pm \infty$ (both) at some asymptotes; and (2) takes the value 1. $\endgroup$
    – David
    Dec 27, 2015 at 15:22
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    $\begingroup$ That idea would be based on the intermediate value theorem and the extreme value theorem. Also, you can simplify things by performing long division of the numerator by the denominator first. $\endgroup$
    – David
    Dec 27, 2015 at 15:23
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    $\begingroup$ You also need to think about the case of $0/0$. $\endgroup$
    – David
    Dec 27, 2015 at 15:28
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    $\begingroup$ Yes, I know, but you would use the intermediate value theorem on intervals not including the poles. Say these are $e$ and $f$. It would be sufficient for the limit to be $+\infty$ to the left of $e$ and $-infty$ to the right of $f$ for instance. I'm realizing it may be a bit more complicated than I said, though, if the "outer" limits are both $+\infty$ or both $-\infty$. What I stated before is not entirely correct. $\endgroup$
    – David
    Dec 27, 2015 at 15:33
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    $\begingroup$ Probably, I should have said "a good deal" more complicated. When there are two vertical asymptotes, there is a lot of checking involved. In that case, the domain is divided into three intervals, and it might be necessary to find the min/max on the inner interval. I was most likely mistaken to believe that this method would be simpler than yours. To complete your proof, besides checking the minor cases I mentioned, all that's left to do is to prove that $a^3 - 9a + 12 \leq 0$ implies $4a^3 + 39 < 0$. $\endgroup$
    – David
    Dec 27, 2015 at 15:45

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I would argue that $4a^3+39<0$ is wrong, for example $a=-3$ satisfies it, however the function, that follows from that value for $a$, does not span all real numbers.

First note that both the numerator and the denominator of the given function are polynomials of order two, with the same coefficients, which means that both limits of $x$ to plus or minus infinity will go to 1. So in order for the function to go to $\pm\infty$ the denominator has to become zero, so the poles of the function have to have real solutions

$$ x=\frac{-1\pm\sqrt{1-4a}}{2}\to a\leq\frac{1}{4}. $$

In order for the function to be equal to zero the numerator has to become zero, so the zeros of the function have to have real solutions

$$ x=\frac{-a\pm\sqrt{a^2-12}}{2}\to a^2\geq 12. $$

Combining these two constraints for $a$ yields

$$ a\leq-2\sqrt{3}. $$

This does not yet ensure that the function covers all real values, namely one of the zeros has to lie between the two poles, such that the function will span all real numbers between the two poles (it will either go from $-\infty$ to $\infty$ or the other way around). Solving this does indeed yield $a^3-9a+12\leq 0$.

So your solution is correct and the given answer (partially) incorrect. Namely when your implicit solution is true, then the given implicit relation is also true. Because for real values of $a$, then your solution can also be approximated as $a\leq-3.5223$ and the given relation as $a<-2.1363$.

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    $\begingroup$ The question appears to have been phrased in the form of an implication, not an equivalence. Your argument doesn't contradict the statement of the question. $\endgroup$
    – David
    Dec 27, 2015 at 16:58
  • $\begingroup$ @fibonatic How do you conclude that the function goes to both $+\infty$ and $-\infty$ at the poles? Might they not go to $+\infty$ at both poles? (when $a= {1 \over 4}$) $\endgroup$
    – Aritra Das
    Dec 27, 2015 at 18:01
  • $\begingroup$ @AritraDas Between the two poles the denominator will be negative, so if the numerator stays positive over that interval, then the function will come from and go to $-\infty$ at that interval between the poles. However if the numerator changes sign in that interval (also crossing zero) then one of the two infinities will then become positive. $\endgroup$ Dec 27, 2015 at 18:24
  • $\begingroup$ I get it now. Since at poles the function changes sign without giong through 0 and at zeroes it changes sign by going through 0, the sufficient condition to have a continuous stretch of $-\infty, 0, \infty$ is to have a zero between 2 poles, in this case. Thanks. $\endgroup$
    – Aritra Das
    Dec 27, 2015 at 18:45

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