46311 與三角形高有關的兩個幾何不等式

\begin{align*} \because\ &S=\frac 12bh_b=\frac 12 ch_c=rp\qquad \therefore\ h_b=\frac{2rp}{b},\quad \ h_c=\frac{2rp}{c},\\ \because\ &abc=4Rrp\\ \therefore\ &\frac{a^4}{h_b^2+h_c^2}=\frac{a^4b^2c^2}{4r^2p^2(b^2+c^2)}=\frac{a^2(abc)^2}{4r^2p^2(b^2+c^2)} =\frac{a^2(4Rrp)^2}{4r^2p^2(b^2+c^2)}=\frac{4a^2R^2}{b^2+c^2}.\\ \because\ &\sum(b^2+c^2)\sum\frac 1{b^2+c^2}\ge 9,\\ \therefore\ &\sum \frac{a^4}{h_b^2+h_c^2}=4R^2\sum \frac{a^2}{b^2+c^2}=4R^2\sum\Big(\frac{a^2+b^2+c^2}{b^2+c^2}-1\Big)\\ &=4R^2\sum\frac {a^2+b^2+c^2}{b^2+c^2}-12R^2\\ &=2R^2\sum(b^2+c^2)\sum\frac 1{b^2+c^2}-12R^2\\ &\ge 2R^2\times 9-12R^2=6R^2.\\ \because\ & b^2+c^2\ge 2bc,\qquad \sum \frac 1{bc}=\frac 1{2Rr},\\ &\qquad \sum a^3=2p(p^2-6Rr-3r^2). \end{align*}

\begin{align*} \therefore\ \sum \frac{a^4}{h_b^2+h_c^2}=\,&4R^2\sum \frac{a^2}{b^2+c^2}\le 4R^2\sum\frac{a^2}{2bc}\\ =\,&2R^2\sum\frac{a^3}{abc}=2R^2\sum\frac{a^3}{4Rrp}=\frac{R\sum a^3}{2rp}\\ =\,&\frac{R\cdot 2p(p^2-6Rr-3r^2)}{2rp}=\frac{R(p^2-6Rr-3r^2)}r\\ \le\, &\frac{R(4R^2+4Rr+3r^2-6Rr-3r^2)}{r}=\frac{2R^2(2R-r)}r,\\ \hbox{故}\quad 6R^2\le &\sum\frac{a^4}{h_b^2+h_c^2}\le \frac{2R^2(2R-r)}{r}. \end{align*}

\begin{align*} &16Rr-5r^2\le p^2\le 4R^2+4Rr+3r^2,\\ \therefore\quad &\frac{6r(2R-r)}{R^2}\le \frac{P^2-4Rr-r^2}{R^2}\le \frac{2(2R^2+r^2)}{R^2},\\ \hbox{故}\quad &\frac{6r(2R-r)}{R^2}\le \sum\frac{h_b^2+h_c^2}{a^2}\le \frac{2(2R^2+r^2)}{R^2}. \end{align*}

$$18\Big(\dfrac{r}{R}\Big)^2\le \sum\dfrac{h_b^2+h_c^2}{a^2}\le \dfrac 92.$$

$$\prod \frac{h_b^2+h_c^2}{a^2}\le \left[\dfrac{\sum\dfrac{h_b^2+h_c^2}{a^2}}{3}\right]^3=\frac 1{27}\Big(\sum \dfrac{h_b^2+h_c^2}{a^2}\Big)^3 \le \dfrac{8(2R^2+r^2)^3}{27R^6},$$ $$\prod \frac{a^2}{h_b^2+h_c^2}\cdot \prod \frac{h_b^2+h_c^2}{a^2}=1,$$

$$\prod \dfrac{a^2}{h_b^2+h_c^2}\ge \dfrac{27R^6}{8(2R^2+r^2)^3}.$$

$$\prod \dfrac{a^2}{h_b^2+h_c^2}\ge \dfrac{8}{27}.$$