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Last edited by hbghlyj 2021-4-16 17:25四边形ABCD内接于圆O,R为其半径.$I_1,I_2$为ΔACD,BCD的内心,$r_1,r_2$为其半径,$d_1=OI_1,d_2=OI_2$,则\[\frac{d_1^2 – d_2^2}{4 (r_2 – r_1)}> r_1 – 2 \sqrt{2 d_1 r_2} \]由欧拉公式有$R=\sqrt{d_1^2+r_1^2}+r_1=\sqrt{d_2^2+r_2^2}+r_2$于是转化为代数题:
正数$r_1,r_2,d_1,d_2$满足$\sqrt{d_1^2+r_1^2}+r_1=\sqrt{d_2^2+r_2^2}+r_2$,求证\[\frac{d_1^2 – d_2^2}{4 (r_2 – r_1)}> r_1 – 2 \sqrt{2 d_1 r_2} \]
MMA验证:
FindInstance[(d1^2 – d2^2)/(4 (r2 – r1)) <= r1 – 2 Sqrt[2 d1 r2] &&
Sqrt[d1^2 + r1^2] + r1 == Sqrt[d2^2 + r2^2] + r2 && d1 > 0 &&
D2 > 0 && r1 > 0 && r2 > 0, {r1, r2, d1, d2}]
输出{},这说明不等式是成立的.
如何证明呢?
一个尝试:若$r_1<r_2$则$d_1>d_2$,有$r_1 – 2 \sqrt{2 d_1 r_2} <r_2 – 2 \sqrt{2 d_2 r_1} $,只需证$\frac{d_1^2 – d_2^2}{4 (r_2 – r_1)}> r_2 – 2 \sqrt{2 d_2 r_1} $.此时交换$r_1,r_2$即化为$r_1>r_2$的情况,故只需考虑$r_1>r_2$的情况,此时$d_1<d_2$,故LHS>0,只需证RHS>0的情形,即$r_1^2>8d_1r_2$.我们要证\[d_2^2 – d_1^2>4( r_1 – 2 \sqrt{2 d_1 r_2} ) (r_1- r_2)\]由2#只需证\[ d_1^3＜128 r_2(r_1- r_2)^2\]但是这个是不成立的 ,例如$r_1= 12,r_2= \frac{1023 \sqrt{145}+12300}{2048},d_1=1,d_2= \frac{1}{32}$ |
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