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kuing 发表于 2024-6-22 16:10 \begin{gather*} (3x+y)^2\leqslant(x^2+3y^2)\left(9+\frac13\right)=\frac{28}3,\\ (3x+y)^2>3x^2+y^2>\f ...
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2025-5-31 10:41 GMT+8
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