特定约束下证明f₁(p,h)大于f₂(p,h)及反区间求解问询

First, let's break down your problem clearly:

• We have two functions:
  $$f_1(p, h) = \frac{1}{p\left(1 - (1-h)^{17+16c_0}\right)}, \quad f_2(p, h) = \frac{1}{ph}$$
• Constraints: $0 < p \leq 1$, $0 < h < 1$, and constant $c_0 > 1$ (so the exponent $k = 17+16c_0 > 33$, which is always greater than 1)

Your intuition suggests $f_1 > f_2$, but let's verify this by manipulating the inequality directly (all denominators are positive here, so we can safely work with them to avoid reciprocal complexity initially).

Step 1: Rewrite the inequality #

For positive values $A$ and $B$, $\frac{1}{A} > \frac{1}{B}$ if and only if $A < B$. Applying this rule to our functions:
$$f_1 > f_2 \iff p\left(1 - (1-h)^k\right) < ph$$

Since $p > 0$, we can divide both sides by $p$ to simplify things further:
$$1 - (1-h)^k < h$$

Rearranging terms, this becomes:
$$(1-h)^k > 1 - h$$

Step 2: Analyze the simplified inequality #

Now we just need to check if $(1-h)^k > 1 - h$ holds for $0 < h < 1$ and $k > 1$.

Let $x = 1 - h$: since $0 < h < 1$, we have $0 < x < 1$. The inequality translates to:
$$x^k > x$$

But for any $0 < x < 1$ and exponent $k > 1$, $x^k < x$ always holds. Here's the quick reasoning:

• $x^k = x \cdot x^{k-1}$
• Since $0 < x < 1$ and $k-1 > 0$, $x^{k-1} < 1$ (raising a number between 0 and 1 to a positive power makes it smaller)
• Multiply both sides by $x$ (positive, so inequality direction stays the same): $x \cdot x^{k-1} < x \cdot 1$, so $x^k < x$

Translating back to $h$, this means:
$$(1-h)^k < 1 - h \implies 1 - (1-h)^k > h$$

Step 3: Final conclusion #

Going back to the original functions:

• $1 - (1-h)^k > h$ implies $p\left(1 - (1-h)^k\right) > ph$ (since $p > 0$)
• Taking reciprocals (both sides are positive, so inequality flips):
  $$\frac{1}{p\left(1 - (1-h)^k\right)} < \frac{1}{ph} \implies f_1 < f_2$$

This holds for all values of $p$ and $h$ within your given constraints, and for any $c_0 > 1$ (since $k > 33 > 1$).

Your initial intuition was incorrect: there is no interval where $f_1 > f_2$—$f_2$ is strictly greater than $f_1$ everywhere in the defined domain.

内容的提问来源于stack exchange，提问作者SpiderRico