> 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
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 <email@example.com> Please feel free to quote anything I say here.