This is exercise 18. d) from Spivak Calculus. Some things I have proven from previous problems that I was using to help me figure this out are:

$x^2+y^2\geq 2xy$; equality when x and y are both 0


I believe I know the answer however I don't think my method is leading to a valid proof.

Here is my process, I rearranged:


$\Rightarrow x^2+y^2>-axy$

$\Rightarrow (x^2+y^2)/xy>-a$

I then considered what value -a, must be in order for this inequality to not hold:

So I want to find where $(x^2+y^2)/xy<-a$

Suppose $xy\geq0$

$x^2+y^2\geq2xy$ $\Rightarrow (x^2+y^2)/xy\geq2$

Thus $-a>(x^2+y^2)/xy\geq2$

Therefore if $(x^2+y^2)/xy<-a$ then $-a>2$. It follows that if xy<0 that -a<-2.

This is about where I'm stuck. I know I can't simply say that since $(x^2+y^2)/xy>-a$ is not true when $-a>2$ or $-a<-2$. That I then know the interval where it is true is $-2<-a<2$.

There's some other odd things I thought about but I'm not sure if they go anywhere. I've spent some time thinking about how x^2+y^2 changes compared to $axy$, how many counterexamples there are when a>2. I also had proven $x^2+xy+y^2>0$ as the sum of 2 squares previously but I don't think it helps me here.

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    $\begingroup$ Try completing the square on $x^2+axy+y^2$ $\endgroup$ – sharding4 Aug 5 '17 at 20:48
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    $\begingroup$ @sharding4 Wow, was much simpler then I was making it out to be. $\endgroup$ – AColoredReptile Aug 5 '17 at 22:28

Suppose $y\ne 0$. We want that $$\forall x\in \Bbb R \;\; f (x)=x^2+axy+y^2>0.$$

with $f (0)=y^2>0$.

the discriminant must be $<0$.

$$\Delta=y^2 (a^2-4)<0$$ which gives

$$-2 <a <2$$

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    $\begingroup$ Problem with your polynomial? OP has $x^2+axy+y^2$ $\endgroup$ – sharding4 Aug 5 '17 at 20:54
  • $\begingroup$ Except the typo with $a\mapsto 2a$, this is much cleaner. Well done (I took the liberty to fix it ) $\endgroup$ – Paolo Leonetti Aug 5 '17 at 20:56
  • $\begingroup$ @PaoloLeonetti Thanks . feel free. $\endgroup$ – hamam_Abdallah Aug 5 '17 at 21:01

Assume $(x,y) \neq (0,0)$ and consider that $$ x^2+axy+y^2=\frac{(2x+ay)^2+(4-a^2)y^2}{4}. $$ This is always $>0$ if: (i) $(4-a^2) \ge 0$. (ii) $(2x+ay)$ and $y$ are not both $0$.

Now (i) means $|a| \le 2$. At this point: if $|a|=2$ then you have $(x\pm y)^2>0$ which is false in general. Otherwise $|a|<2$: here, $2x+ay=y=0$ if and only if $x=y=0$, which has been excluded by hypothesis.

Hence, the answer is $a \in (-2,2)$.


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