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How
to Write a Solution
sdrawkcaB knihT, Write Forwards
by
Mathew Crawford
The following is an excerpt from a cookbook that was never written:
"Figuring out how to make an omelette is easy. Anybody who has eaten an omelette knows that an omelette is typically made with several eggs filled with various foods such as ham, peppers, onions, and bacon and is often cooked with cheese. The fact that all these ingredients end up inside the egg means that we should begin cooking the eggs flatly on a pan and then add the ingredients. We can then roll part of the egg over the ingredients so as to trap them on the inside. If we needed some of the ingredients precooked we could do that before adding them to the eggs..."
It is one thing to figure out how to make an omelette. It is another to explain to somebody else how to make one. Starting our explanation from the beginning is much clearer than starting with the finished omelette.
"Prepare vegetables and other desired omelette fillings. Beat eggs. Start cooking the eggs. Add your fillings in the middle so that part of the egg can be pulled over the ingredients. When the omelette is closed, continue to cook and flip the omelette until the eggs look well-cooked."
The reader doesn't care how the process of cooking an omelette was unraveled by the author. The reader just wants to know how to make an omelette.
Think of solutions as recipes. Start at the beginning and move forward. List the ingredients and explain how and when to add them to the pot.
Here's a sample problem:
3abc.
This solution might be a good way to see how we might come up with a solution from scratch, but it's not a particularly well-written proof:
We
note that the inequality contains the factors (-a + b + c), (a - b +
c), and
(a + b - c). These factors point to using the triangle inequality so
it seems natural to leave the factors alone and invoke the fact that
each is nonnegative.
Since each of these three factors is multiplied by the square of the length of a side it might be possible to manipulate the inequality into something involving these nonnegative triangle inequality factors multiplied by perfect squares. We could then argue that this sum must also be nonnegative. We begin by moving 3abc to the left hand side:
a2(-a
+ b + c) + b2(a - b + c) + c2(a + b - c) - 3abc
0.
If we were to view 3abc as the sum of 3 terms that are each the product of ab, bc, or ca and one of the triangle inequality factors, we begin to get an idea as to how the inequality can be reorganized. Since the inequality is cyclic, it seems natural to take these products in a way that preserves the cyclic nature. For instance, we multiply ab with (a + b - c) because a and b have the same sign in (a + b - c):
ab(a + b -
c) = a2b + ab2 - abc 
0,
bc(-a + b + c) = -abc + b2c + bc2
0,
ca(a - b + c) = a2c - abc + ac2
0.
We see the -3abc
in the sum of these products. Examining the other terms in
ab(a + b - c) = a2b + ab2 - abc, we notice that
a2b + ab2 are factors that would pop out of (a
- b)2(a + b - c) . Expanding the squared parts of expressions
like (a - b)2(a + b - c) we get
(a - b)2(a
+ b - c) = a2(a + b - c) - 2ab(a + b - c) + b2(a
+ b - c)
0,
(b - c)2(-a + b + c) = b2(-a + b + c) - 2bc(-a
+ b + c) + c2(-a + b + c)
0,
(c - a)2(a - b + c) = c2(a - b + c) - 2ca(a -
b + c) + a2(a - b + c)
0.
Adding these inequalities together we begin to see the inequality take shape:
a2(a
+ b - c + a - b + c) + b2(a + b - c - a + b + c) + c2(-a
+ b + c + a - b + c) +
- 2a2b - 2ab2 + 2abc + 2abc - 2b2c
- 2bc2 - a2c + abc - ac2 =
a2(2a - 2b - 2c) + b2(-2a + 2b - 2c) + c2(-2a
- 2b + 2c) + 6abc
0.
Muliplying this inequality by -1/2 reversed the inequality sign and gives us
a2(-a
+ b + c) + b2(a - b + c) + c2(a + b - c) - 3abc
0.
Adding 3abc to both sides gives us
a2(-a
+ b + c) + b2(a - b + c) + c2(a + b - c)
3abc
and we are done.
The cookbook style is easier to read and far more convincing:
According to the triangle inequality, the sum of any two sides of a triangle is at least as great as the length of the third side. Thus, we have the inequalities
a
- b - c 0,
b - c - a
0,
c - a - b
0.
Multiplying these by perfect squares leave each left-hand side nonpositive, so
(b
- c)2(a - b - c)
0,
(c - a)2(b
- c - a) 0,
(a - b)2(c
- a - b) 0,
or
(b2
- 2bc + c2)(a - b - c)
0,
(c2
- 2ac + a2)(b - c - a)
0,
(a2
- 2ab + b2)(c - a - b)
0,
Adding these inequalities we get
a2(b
- c - a + c - a - b) + b2(a - b - c + c - a - b) + c2(a
- b - c + b - c - a)
- 2bc(a -
b - c) - 2ca(b - c - a) - 2ab(c - a - b) =
= 2a2(-a
+ b + c) + 2b2(a - b + c) + 2c2(a + b - c) - 6abc
0.
Adding 6abc to both sides and dividing by 2 we have the desired
a2(-a
+ b + c) + b2(a - b + c) + c2(a + b - c)
3abc.
