不等式吧 关注：9,733贴子：51,782
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\[\begin{array}{l}
\Leftrightarrow \frac{{3{\rm{ + }}2{p^2} - 4q + {q^2} - 2p}}{{2 + {q^2} - 2p + {p^2} - 2q}} + \frac{8}{{{{\left( {1 + p} \right)}^2}}} \le 2\\
\Leftrightarrow f\left( q \right) = \left( {2 - \frac{8}{{{{\left( {1 + p} \right)}^2}}}} \right)\left( {2 + {q^2} - 2p + {p^2} - 2q} \right) - \left( {3{\rm{ + }}2{p^2} - 4q + {q^2} - 2p} \right) \ge 0\\
f'\left( q \right) = 4\left( {q - 1} \right)\left( { - \frac{4}{{{{\left( {1 + p} \right)}^2}}}} \right) \le 0
\end{array}\]
\Leftrightarrow \frac{{3{\rm{ + }}2{p^2} - 4q + {q^2} - 2p}}{{2 + {q^2} - 2p + {p^2} - 2q}} + \frac{8}{{{{\left( {1 + p} \right)}^2}}} \le 2\\
\Leftrightarrow f\left( q \right) = \left( {2 - \frac{8}{{{{\left( {1 + p} \right)}^2}}}} \right)\left( {2 + {q^2} - 2p + {p^2} - 2q} \right) - \left( {3{\rm{ + }}2{p^2} - 4q + {q^2} - 2p} \right) \ge 0\\
f'\left( q \right) = 4\left( {q - 1} \right)\left( { - \frac{4}{{{{\left( {1 + p} \right)}^2}}}} \right) \le 0
\end{array}\]
