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[不等式] 不等式题(似乎可以与函数关联)

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星奔川骛 Posted at ereyesterday 23:15 |Read mode
$$
x \in (0, 1) \quad y \in \mathbb{R} \quad \text{证明:}
$$

$$
x^y + (1 - x)^{2 - y} \geq 1 - y^2 (2 - y)^2 \left( \frac{1}{16} - x^2 (1 - x)^2 \right)
$$

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2025-4-20 12:09 GMT+8

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