46411 三角形中線與高之間的兩個幾何不等式

$$\frac{m_a}{h_a}\ge \frac{(b+c)^2}{4bc}, \ \hbox{僅當}\ b=c\ \hbox{時等號成立。}$$

$\therefore$ $\sum \dfrac 1a=\dfrac{\sum ab}{\prod a}=\dfrac{p^2+4Rr+r^2}{4Rrp}$.

$\because$ $m_a^2=\dfrac 14(2b^2+2c^2-a^2)$, $\sum a^2=2(p^2-4Rr-r^2)$,

\begin{align*} \therefore\quad \dfrac{\sum m_a^2}{\sum h_a}=&\dfrac{\sum \dfrac 14(2b^2+2c^2-a^2)}{\sum \dfrac{2rp}{a}}=\dfrac{\dfrac 14\sum (2b^2+2c^2-a^2)}{2rp\sum \dfrac{1}{a}}\\ =&\dfrac{\dfrac 34\sum a^2}{2rp\cdot\dfrac{p^2+4Rr+r^2}{4Rrp}}=\dfrac{\dfrac 32 R\sum a^2}{p^2+4Rr+r^2}=\dfrac{3R(p^2-4Rr-r^2)}{p^2+4Rr+r^2}\\ =&\dfrac{3R(p^2+4Rr+r^2-8Rr-2r^2)}{p^2+4Rr+r^2}=3R-\dfrac{3R(8Rr+2r^2)}{p^2+4Rr+r^2}. \end{align*}

$$16Rr-5r^2\le p^2\le 4R^2+4Rr+3r^2,$$ $$\therefore\quad 3R-\frac{3R(8Rr+2r^2)}{(16Rr-5r^2)+4Rr+r^2}\le \dfrac{\sum m_a^2}{\sum h_a}\le 3R-\frac{3R(8Rr+2r^2)}{(4R^2+4Rr+3r^2 )+4Rr+r^2};$$

$$\hbox{於是 ：}\quad \frac{9R(2R\!-\!r)}{2(5R\!-\!r)} =\frac{9R[(5R\!-\!r)\!-\!3R]}{2(5R\!-\!r)} =\dfrac 92 R\!-\!\dfrac{27R^2}{2(5R\!-\!r)} \ge \dfrac 92 R\!-\!\dfrac{27R^2}{2(5R\!-\!\frac R2)} =\dfrac 32 R.$$