Inequalities

Hint for third inequality:

Using $a^2+b^2\ge 2ab$ we get $(a+b)^2=a^2+b^2+2ab\ge 2ab+2ab=4ab$. So square rooting both sides (since $a,b\ge 0$) we get $a+b\ge 2\sqrt{ab}$, i.e. $(a+b)(b+c)(c+a)\ge 2\sqrt{ab}(b+c)(c+a)$. Do you see how to continue?

Hint for first inequality:
Clearly we must either have:

1) x+y>0 and x+2y-5<0 or
2) x+y<0 and x+2y-5>0.

For case 1 we can "solve" the first inequality to get $x>-y$, so putting $x=-y+a$ where $a>0$ gives:

$(a+y-5)(a)<0$.

So again either:
Case 1.1) a+y-5<0 and a>0 or
Case 1.2) a+y-5>0 and a<0

etc.

I know this isn't the best way.

Just to simplify the factors. Because since $x>-y$ in the case we were considering, we can just set $x=-y+a$ where a>0.

To solve for x,y you can consider the sub-cases 1.1 and 1.2.

In 1.1 we assume that a+y-5<0 and a>0.

From a+y-5<0 we get a<5-y, and since a>0 also holds, we have:

0<a<5-y.

Since x=-y+a by definition, we have:

-y+0<x=-y+a<-y+5-y, or equivalently,

-y<x<5-2y

Testing we see that this gives a valid inequality for case 1, so this is a solution.

EDIT: Do the same thing for cases 1.2 and 2.

Case 2 is entirely different from case 1. I.e. the "case tree" should look something like this:

Case 1)
-Case 1.1)
-Case 1.2)

Case 2)
-...

You can test them by considering what happens if 5-2y<x<-y.

The inequality is (x+(2y-5))(x+y)<0

Now x>5-2y implies x+(2y-5)>0, and x<-y implies x+y<0, so we have:

x+(2y-5)=+ve and x+y=-ve, hence

(x+(2y-5))(x+y)=(+ve)(-ve)=-ve<0, as desired.

So indeed 5-2y<x<-y is a solution. But you need to remember to state that this only holds for y>5 (I forgot to do this in my last post). The one I did (i.e. case 1.1) only holds for y<5. And for y=5 we get $(x+5)^2<0$ which is impossible by the trivial inequality.

1.1 only holds when y<5, because -y<x<5-2y implies that -y<5-2y, i.e. y<5. Similarly for 1.2.

Case 2 is defined in my first hint:


BTW, I just realized that I over-complicated it for nothing. Here is an easy approach which anyone can understand. Can't believe I just found it.

Notice that if (x+2y-5)(x+y)<0 then either:

Case 1) x+2y-5<0 and x+y>0, or

Case 2) x+2y-5>0 and x+y<0.

In case 1, the inequality x+2y-5<0 implies that x<5-2y. Similarly, x+y>0 implies that x>-y. Therefore, for Case 1 to hold we must have -y<x<5-2y, indeed a solution. (You can check using the way I showed above.)

Now do a similar thing for case 2.

Yes, sorry. And the rest of your working is correct. The second inequality shouldn't be too bad now. I think the tricky part was figuring out how you were meant to state the solution, since there are two variables.

Yes, sorry. And the rest of your working is correct. The second inequality shouldn't be too bad now. I think the tricky part was figuring out how you were meant to state the solution, since there are two variables.

You're right about case 2 being absurd.

Hint for case 1: Think geometrically (you anticipated this in your opening post). For example, what does x^2+y^2=9 represent geometrically? Now extend this line of thought to 4<x^2+y^2<9.

In fact, the geometric approach also provides a neat approach to the first inequality. But you have lines instead of

i.e.

(x+2y-5)(x+y)<0

Case 1) x+2y-5<0 and x+y>0, so:
x+2y<5 and x+y>0 (*)
Now consider the lines x+2y=5 and x+y=0. (The inequalities (*) might be easier if you write them in "y=mx+c" form, i.e. $y<-\frac{1}{2} x+\frac{5}{2}$, etc.)

Case 2 is similar.

I prefer the "analytic" approach to the geometric because it generalizes more easily (you can't always sketch a diagram for an inequality, or at least it won't always be obvious how it looks like).

..

Hint: $4<x^2+y^2<9$ implies that $y^2<9-x^2$ AND $y^2>4-x^2$. Solving these we get:

1) $y<\sqrt{9-x^2}$ OR $y>-\sqrt{9-x^2}$

AND

2) $y>\sqrt{4-x^2}$ OR $y<-\sqrt{4-x^2}$

In other words, you can take any of the two inequalities in (1) and combine them with any of the two inequalities in (2) and see what that gives you.

So for example taking $y<\sqrt{9-x^2}$ from (1) and $y>\sqrt{4-x^2}$ from (2), we get:

$\sqrt{4-x^2}<y<\sqrt{9-x^2}$.

EDIT: And keep in mind that in order for $\sqrt{4-x^2}$ to be real, we need $4-x^2\ge 0$ or $x^2\ge 4$ or $-2\le x\le 2$.

We're somewhat guessing at the actual intent of the question given the lack of context, but I don't really like what you've done above (plus it actually has errors):

At university level, x^2+y^2 < 9 should be automatically recognized as the interior of a circle around the origin of radius 3.

Rewriting as "$y < \sqrt{9-x^2}$ OR $y > -\sqrt{9-x^2}$" does not to my mind move things forwards - it just confuses matters.

Unfortunately, that confusion is shown by the fact that it's not actuially correct: as long as 9-x^2 > 0, then for all values of y, it is always the case that one of $y < \sqrt{9-x^2}$, $y > -\sqrt{9-x^2}$" is true.

It's an easy mistake to make, but that's my point - you're moving things into a complexity of having to consider square root signs, whether or not quantities are real, etc, when the original x^2+y^2<9 equation avoids all of this.

[Mileage may vary on this, depending on how "obvious" you find what x^2+y^2 < 9 represents].

Not at all valid. As you've said, the area you want is the "ring shape" between the circles. Technically, it is an annulus of inner radius 2 and outer radius 3 centered at the origin.

Any analytic expression you try to form to describe the region is going to be horrific.

For example, the upper quadrant of the region would need to be treated as two separate regions:

The part of the region where 0 <=x<=2 can be written as: $0 \le x \le 2, y \ge 0, x \le 2, \sqrt{4-x^2} < y < \sqrt{9-x^2}$.
The part of the region where x > 2 can be written as $2 < x < 3, 0 \le y < \sqrt{9-x^2}$

You then have the 3 other quadrants to describe.

To my mind, doing this is completely pointless and misses the point (unless, for example, you need to fill the region by writing a computer program, when it just *might* be the correct approach).

Putting the mistake to one side (thanks for pointing that out!), I was trying to solve it in terms of x or y, i.e. "analytically", because you can't just refer to the diagram (and I did refer to the diagram -- see post #17) in more complicated inequalities. For example, solve $(x^4+y^2-4)(y^4+x^2-9)<0$ for $x,y\in \mathbb{R}$.