# Inequality problem

1. Jul 11, 2013

### Saitama

1. The problem statement, all variables and given/known data
Let $x^2+y^2+xy+1 \geq a(x+y)$ for all $x,y \in R$. Find the possible integer(s) in the range of $a$.

2. Relevant equations

3. The attempt at a solution
I can rewrite this into $(x+y)^2-xy+1 \geq a(x+y) \Rightarrow (x+y)(x+y-a)+1-xy \geq 0$ but I don't think that this would help.

2. Jul 11, 2013

### szynkasz

You can treat this as a quadratic inequality with a variable $x$:

$x^2+(y-a)x+y^2-ay+1\ge 0$

and determine when this parabola takes only non-negative values.

3. Jul 11, 2013

### verty

I don't have any great advice but the method of Lagrange multipliers (from calculus) shows that the = case happens when x = y. Probably one must by some argument show that the strongest constraining effect happens when x = y. Or perhaps not, but I don't have any better advice.

Last edited: Jul 11, 2013
4. Jul 11, 2013

### verty

I've got it. You must rotate the coordinate axes. Rotate them 45 degrees counter-clockwise, this will make the right side depend on a single variable. I was looking for a way to make this problem easier.

$(x,y) → (\frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}}, \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}})$

5. Jul 11, 2013

### haruspex

It doesn't need to be a pure rotation, so you can forget the √2s. Any invertible substitution is valid.

6. Jul 12, 2013

### Alpharup

You have already found one equation;
$(x+y)^2-xy+1 \geq a(x+y)$
This can again be rewritten as $(x+1)^2+(y+1)^2-2(x+y)+xy-1\geq a(x+y)$

Add these two equations and with little manupulation you get,

$(x+1)^2+(y+1)^2+(x+y)^2 \geq 2(a+1)(x+y)$

Then with some more manupulation you get,

$-(x+y)(x+y-2a-2) \geq (x+1)^2+(y+1)^2$ Divide by 2..

$(-(x+y)(x+y-2a-2))/2 \geq ((x+1)^2+(y+1)^2)/2$

here $(x+1)^2$ and $(y+1)^2$ are both positive. So, their arithmetic mean is ≥ geometric mean
which implies,

$(-(x+y)(x+y-2a-2))/2 \geq sqrt(x+1)(y+1)$

I think we have got nearer to the answer. Hope this helps

Last edited: Jul 12, 2013
7. Jul 12, 2013

### verty

Right, I didn't see that. So any invertible substitution that takes x+y to a single-variable alternative will work to make the problem solvable without calculus or advanced techniques. The basic idea is to let a depend on x only, so we can set y to 0 without affecting anything.

8. Jul 13, 2013

### Saitama

Sorry for such a late reply.

The discriminant should be less than zero i.e.
$$(y-a)^2-4(y^2-ay+1)<0$$
What am I supposed to do with this? Work on the discriminant of the resulting quadratic again?

Is it okay to add them? :uhh:

9. Jul 13, 2013

### szynkasz

The discriminant should be less or equal zero as the value must be non-negative, not positive. And yes, work on the discriminant again.

10. Jul 13, 2013

### Alpharup

If a<b and c<d
then a+b<c+d..

11. Jul 14, 2013

### Saitama

Thank you! That is a nice method.

12. Jul 14, 2013

### szynkasz

You are welcome