求证：当x>0、y>0时，log(1+x/y) ≤ x/y是否成立？

Hey there! Let's break this down clearly and directly—great question, by the way.

First, let's simplify the problem with a substitution: let ( t = \frac{x}{y} ). Since ( x > 0 ) and ( y > 0 ), this means ( t > 0 ). Now our question boils down to: does ( \log(1+t) \leq t ) hold for all ( t > 0 )?

Proof 1: Using Calculus (Function Monotonicity) #

Let's define a function ( f(t) = t - \log(1+t) ). Here's how we analyze it:

1. Calculate the derivative: ( f'(t) = 1 - \frac{1}{1+t} = \frac{t}{1+t} ).
2. For ( t > 0 ), both the numerator and denominator are positive, so ( f'(t) > 0 ). This means ( f(t) ) is strictly increasing on the interval ( (0, +\infty) ).
3. Evaluate ( f(t) ) at ( t=0 ): ( f(0) = 0 - \log(1+0) = 0 ).

Since ( f(t) ) is strictly increasing and starts at 0 when ( t=0 ), for all ( t > 0 ), ( f(t) > f(0) = 0 ). Rearranging this gives ( t - \log(1+t) > 0 ), so ( \log(1+t) < t ). Obviously, this satisfies the weaker condition ( \log(1+t) \leq t ).

Proof 2: Using Taylor Series Expansion #

For ( t > -1 ), the natural logarithm has the alternating Taylor series expansion:
[
\log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \dots
]
When ( t > 0 ), each term after the first alternates sign and gets smaller in absolute value. The remainder after the first term is a negative value (since we subtract a larger positive term before adding a smaller one), so:
[
\log(1+t) = t - (\text{positive value}) < t
]
This again confirms the inequality holds.

Quick Note on Logarithm Base #

If you're using base-10 logs instead of natural logs, the result still stands. Since ( \log_{10}(1+t) = \frac{\ln(1+t)}{\ln(10)} ), and ( \ln(10) \approx 2.302 > 1 ), we get:
[
\log_{10}(1+t) = \frac{\ln(1+t)}{\ln(10)} < \frac{t}{\ln(10)} < t
]
So the inequality is true no matter which standard positive base (>1) you use for the logarithm.

内容的提问来源于stack exchange，提问作者Ramin