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Topic: Second to last algebra question
Replies: 1   Last Post: May 2, 2005 11:42 AM

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Vincent Johns

Posts: 89
Registered: 1/25/05
Re: Second to last algebra question
Posted: May 2, 2005 11:42 AM
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Original problem (as corrected):

> Prove that a^2 + b^2 + 1 >= a + b + ab.

elanamig wrote:

> Hi, Brian,
> Worked it out, and it works (as expected). I would have never thought
> to use substitution of variables to solve this. Is there an approach
> that does not require introducing new variables?
> Thanks,
> Elana

Here's another approach:

> Prove that a^2 + b^2 + 1 >= a + b + ab.

This is equivalent to proving that, for all real a and b,

a^2 + b^2 + 1 - a - b - ab >= 0 .

We can think of this as a contour plot, on a rectangular (Cartesian) map
with a and b coordinates. At each point (a, b) on the map, we can look
at the value of the function represented by the left side of this
inequality. For any given values of a and b, let's define

k := a^2 + b^2 + 1 - a - b - ab

(Actually, I'm thinking of k(a, b) as a function with arguments of a and
b, but for simplicity, let's think of it as just a real number.)

If we prove that k must always be non-negative, we're done.

OK. Let's choose a value for b and calculate what a might be in terms
of b and k.

a^2 + b^2 + 1 - a - b - ab = k
a^2 - a*(b + 1) + b^2 - b + 1 - k = 0

It's a ho-hum quadratic equation. Plugging into the formula, we get (up
to) two solutions, and we're interested only in real solutions. Unless
I made a mistake (sorry, I'd use Mathematica if I had a copy, but I did
this by hand), I get

a = ((b + 1) +- sqrt(-3*(b^2 - 2*b + 1)+ 4*k)) / 2
a = ((b + 1) +- sqrt(-3*(b - 1)^2 + 4*k)) / 2

For a to be a real number, the argument of the square-root function must
be non-negative, and the -3 makes that difficult if k is negative. Even
if k = 0, the only way a can be real is for b to equal 1, and then a =
1. (QED, as they used to say. Nowadays, they use a little black
rectangle, which I have a hard time typing.)


All this said, let me make a sales pitch for learning to recognize
patterns like Brian's substitution. Notice that the function is
symmetrical in a and b. Choosing a substitution involving the mean of
the two values is a clever idea, and it paid off.


Another approach that I considered was to think of a Cartesian plot in
(a, b) coordinates, but rotating the axes to remove the (ab) term and
make the remaining terms form a more familiar pattern (such as a conic
section of some sort), whose properties you know. (I didn't actually do
this, but it might not be a bad thing to try.)

-- Vincent Johns <>
Please feel free to quote anything I say here.

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