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[不等式] 三角不等式

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hbghlyj Posted 2020-12-28 21:08 |Read mode
Last edited by hbghlyj 2020-12-28 21:20在三角形ABC中,证明\[\cos^2A+\sin A\cos B\cos C\le\frac{11}{30}+\frac{1}{60} \sqrt{3781} \cos \left(\frac{1}{3} \cos ^{-1}\left(-\frac{38746}{3781 \sqrt{3781}}\right)\right)\]
取等时$B=C=\tan^{-1}\left(\frac{1}{4}+\frac{1}{2} \sqrt{\frac{19}{3}} \cos \left(\frac{1}{3} \left(\cos ^{-1}\left(\frac{3 \sqrt{\frac{3}{19}}}{19}\right)-2 \pi \right)\right)\right)$.

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kuing Posted 2020-12-28 23:15
\[\LHS=\cos^2A+\sin A\cdot\frac{\cos(B-C)+\cos(B+C)}2\leqslant t^2+\sqrt{1-t^2}\cdot\frac{1-t}2,\]其中 `t=\cos A`,然后就变成了求一元函数最大值,结果是取等的 `t` 满足 `20t^3+16t^2-3t-1=0` 以及最大值 `M` 满足 `320M^3-352M^2-123M+91=0`,都是三次方程,解出来就是那吓人的样子了。

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