A few days ago, in the mathematics paper of the Yunfu Grade 11 final exam, there was a problem that left a deep impression on me. At that time, I was unable to prove it formally and only arrived at the answer by using specific examples. I have just given it some more thought and will present the proof process here. The problem is as follows:
A function f(x) defined on (0, +\infty) satisfies x f'(x) \leq f(x). For any 0 < a < b, compare the magnitudes of a f(b) and b f(a).
Since this was a fill-in-the-blank question, I obtained the result by testing two examples:
Let f(x) = x, then a f(b) = a b and b f(a) = b a, so a f(b) = b f(a).
Let f(x) = x + 1, then a f(b) = a(b + 1) = ab + a and b f(a) = b(a + 1) = ba + b. Since a < b, it follows that a f(b) < b f(a).
Thus, the answer is a f(b) \leq b f(a).
Of course, what readers want to know is not just the answer, but how to come up with these examples. During the exam, BoJone was momentarily without a clear direction, so I thought about replacing the inequality x f'(x) \leq f(x) with an equality. Solving the differential equation x f'(x) = f(x) yields f(x) = kx, where k is an arbitrary constant. Substituting this into a f(b) and b f(a) results in an identity. By slightly modifying this example (adding a constant term), I obtained the inequality case. Thus, the problem was solved.
However, truly good mathematics requires rigorous proof. This is what I came up with this evening:
To compare the magnitudes of a f(b) and b f(a), we only need to compare the magnitudes of \frac{a f(b)}{ab} and \frac{b f(a)}{ab}, which is equivalent to comparing \frac{f(b)}{b} and \frac{f(a)}{a}. This effectively means proving the monotonicity of the function \frac{f(x)}{x}.
The derivative is: \left(\frac{f(x)}{x}\right)' = \frac{x f'(x) - f(x)}{x^2}
The problem states that x f'(x) \leq f(x), therefore: \left(\frac{f(x)}{x}\right)' = \frac{x f'(x) - f(x)}{x^2} \leq 0 This implies that \frac{f(x)}{x} is either a decreasing function or a constant function. Given that 0 < a < b, we have: \frac{f(b)}{b} \leq \frac{f(a)}{a} which simplifies to: a f(b) \leq b f(a)
In fact, BoJone believes that this problem should not have been a fill-in-the-blank question but rather a comprehensive problem requiring the full solution process. Such problems truly qualify as “gymnastics for exercising the mind.”
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