# Chương II : Tam giác

8 tháng 9 2017 lúc 8:55

Ta có:

$S=pr=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}$

$\Leftrightarrow p^2r^2=p\left(p-a\right)\left(p-b\right)\left(p-c\right)$

$\Leftrightarrow r^2=\dfrac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}$

$\Leftrightarrow\dfrac{1}{r^2}=\dfrac{p}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\dfrac{1}{\left(p-a\right)\left(p-b\right)}+\dfrac{1}{\left(p-b\right)\left(p-c\right)}+\dfrac{1}{\left(p-c\right)\left(p-a\right)}$

$\Leftrightarrow\dfrac{1}{r^2}=4\left(\dfrac{1}{\left(b+c-a\right)\left(c+a-b\right)}+\dfrac{1}{\left(c+a-b\right)\left(a+b-c\right)}+\dfrac{1}{\left(a+b-c\right)\left(b+c-a\right)}\right)$

$\Leftrightarrow\dfrac{1}{4r^2}=\dfrac{1}{c^2-\left(a-b\right)^2}+\dfrac{1}{a^2-\left(b-c\right)^2}+\dfrac{1}{b^2-\left(c-a\right)^2}\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

$\Leftrightarrow\dfrac{1}{r^2\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}\ge4\left(1\right)$

Ta lại có:

$S=\dfrac{ah_a}{2}=pr=\dfrac{r\left(a+b+c\right)}{2}$

$\Leftrightarrow h_a=\dfrac{r\left(a+b+c\right)}{a}$

$\Leftrightarrow h_a^2=\dfrac{r^2\left(a+b+c\right)^2}{a^2}\left(2\right)$

Tương tự ta có:

$\left\{{}\begin{matrix}h_b^2=\dfrac{r^2\left(a+b+c\right)^2}{b^2}\left(3\right)\\h_c^2=\dfrac{r^2\left(a+b+c\right)^2}{c^2}\left(4\right)\end{matrix}\right.$

Từ (2), (3), (4) ta có:

$h_a^2+h_b^2+h_c^2=r^2\left(a+b+c\right)^2\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$

$\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{h_a^2+h_b^2+h_c^2}=\dfrac{1}{r^2\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}\ge4$

Bình luận (9)
8 tháng 9 2017 lúc 13:47

Kẽ đường thẳng (d) đi qua A và // với BC. Gọi B' đối xứng với B qua (d).

Ta có:

$BB'^2=B'C^2-BC^2\le\left(AB'+AC\right)^2-BC^2$

$\Leftrightarrow4h_a^2\le\left(b+c\right)^2-a^2\left(1\right)$

Tương tự ta cũng có:

$\left\{{}\begin{matrix}4h_b^2\le\left(c+a\right)^2-b^2\left(2\right)\\4h_c^2\le\left(a+b\right)^2-c^2\left(3\right)\end{matrix}\right.$

Cộng (1), (2), (3) vế theo vế ta được

$4h_a^2+4h_b^2+4h_c^2\le\left(a+b\right)^2-c^2+\left(b+c\right)^2-a^2+\left(c+a\right)^2-b^2$

$\Leftrightarrow4\left(h_a^2+h_b^2+h_c^2\right)\le\left(a+b+c\right)^2$

$\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{h_a^2+h_b^2+h_c^2}\ge4$

Bình luận (0)
5 tháng 2 2018 lúc 20:09

sai đề

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