当B取较大值时，给定代数不等式是否成立的严谨验证问询

Let's break this down step by step to rigorously check whether the inequality holds for large values of ( B ), using algebraic manipulation and asymptotic analysis.

First, let's restate the original inequality and known conditions for clarity:

We need to check if:
$$\frac{-AB^{2}{B-x_2}+\frac{2y_2}{B}<\frac{-AB}2}{B-x_1}+\frac{2y_1}{B}$$
Given: ( x_2 > x_1 ), ( B > x_2 ), ( y_2 > y_1 ), and all variables are positive.

Step 1: Simplify the Inequality by Analyzing the Difference #

A common strategy for verifying inequalities is to compute the difference between the left-hand side (LHS) and right-hand side (RHS), then check if this difference is negative. Let's define ( D = \text{LHS} - \text{RHS} ):

$$
D = \left( \frac{-AB^2}{B-x_2} + \frac{2y_2}{B} \right) - \left( \frac{-AB^2}{B-x_1} + \frac{2y_1}{B} \right)
$$

Simplify ( D ) by combining like terms:
$$
D = AB^2 \left( \frac{1}{B-x_1} - \frac{1}{B-x_2} \right) + \frac{2(y_2 - y_1)}{B}
$$

Combine the fractions in the parentheses:
$$
\frac{1}{B-x_1} - \frac{1}{B-x_2} = \frac{(B-x_2) - (B-x_1)}{(B-x_1)(B-x_2)} = \frac{x_1 - x_2}{(B-x_1)(B-x_2)}
$$

Substitute back into ( D ), and note that ( x_1 - x_2 = -(x_2 - x_1) ) (since ( x_2 > x_1 )):
$$
D = -AB^2 \cdot \frac{x_2 - x_1}{(B-x_1)(B-x_2)} + \frac{2(y_2 - y_1)}{B}
$$

Our goal is to show ( D < 0 ) for sufficiently large ( B ).

Step 2: Asymptotic Analysis as ( B \to \infty ) #

Let's examine the behavior of ( D ) when ( B ) becomes very large. For large ( B ), the denominator ( (B-x_1)(B-x_2) \approx B^2 ) (since the lower-order terms ( -(x_1+x_2)B + x_1x_2 ) become negligible compared to ( B^2 )).

Substitute this approximation into ( D ):
$$
D \approx -AB^2 \cdot \frac{x_2 - x_1}{B^2} + \frac{2(y_2 - y_1)}{B} = -A(x_2 - x_1) + \frac{2(y_2 - y_1)}{B}
$$

As ( B \to \infty ), the term ( \frac{2(y_2 - y_1)}{B} \to 0 ), so the limit of ( D ) is:
$$
\lim_{B \to \infty} D = -A(x_2 - x_1)
$$

Since ( A > 0 ) and ( x_2 > x_1 ), this limit is strictly negative. This means that as ( B ) grows arbitrarily large, ( D ) will eventually become and remain negative.

Step 3: Rigorous Verification for Finite Large ( B ) #

To confirm the inequality holds for all sufficiently large finite ( B ), we can rewrite ( D ) using ( t = \frac{1}{B} ) (where ( t \to 0^+ ) as ( B \to \infty )):
$$
D(t) = -\frac{A(x_2 - x_1)}{1 - (x_1+x_2)t + x_1x_2 t^2} + 2(y_2 - y_1)t
$$

Using the Taylor expansion ( \frac{1}{1 - z} \approx 1 + z + z^2 + \dots ) for small ( z ), we expand the denominator:
$$
D(t) \approx -A(x_2 - x_1)\left(1 + (x_1+x_2)t\right) + 2(y_2 - y_1)t
$$
$$
D(t) = -A(x_2 - x_1) + \left[ -A(x_2 - x_1)(x_1+x_2) + 2(y_2 - y_1) \right]t
$$

Since the leading term ( -A(x_2 - x_1) < 0 ), there exists some small ( t_0 > 0 ) (corresponding to ( B_0 = \frac{1}{t_0} > x_2 )) such that for all ( t < t_0 ) (i.e., ( B > B_0 )), ( D(t) < 0 ). This directly implies the original inequality holds for all ( B > B_0 ).

Step 4: Edge Case Check (Near ( B = x_2 )) #

Note that for ( B ) just slightly larger than ( x_2 ), the inequality might not hold (depending on the values of ( A, y_1, y_2 )), but this is irrelevant to your question about large ( B ). As ( B ) increases beyond a certain threshold, the cubic term in the simplified inequality will dominate, ensuring ( D < 0 ).

Conclusion #

For sufficiently large ( B ):

• The difference ( D = \text{LHS} - \text{RHS} ) is strictly negative.
• The original inequality ( \text{LHS} < \text{RHS} ) holds rigorously.

Your initial intuition was correct—while the leading terms of LHS and RHS appear equal as ( B \to \infty ), the next-order terms reveal that LHS is strictly smaller than RHS.

内容的提问来源于stack exchange，提问作者pafnuti