Cho tam giác $ABC$. Chứng minh rằng: ${m_a}{m_b}{m_c} \ge S.p \ge {l_a}{l_b}{l_c}$
Ta có: $4{m_a}^2 = 2\left( {{b^2} + {c^2}} \right) - {a^2}$ $ \ge {\left( {b + c} \right)^2} - {a^2} = \left( {b + c - a} \right)\left( {b + c + a} \right)$
$\Rightarrow m^2_a \ge p(p-a)$
Tương tự: $\left\{ \begin{array}{l}
m^2_b \ge p(p-b)\\
m^2_c \ge p(p-c)
\end{array} \right.$
Suy ra :${m_a}^2{m_b}^2{m_c}^2 \ge {p^2}.p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)$
    $\begin{array}{l}
 \Rightarrow {m_a}^2{m_b}^2{m_c}^2 \ge {p^2}.{S^2}\\
 \Rightarrow {m_a}{m_b}{m_c} \ge pS
\end{array}$
Đẳng thức xảy ra $ \Leftrightarrow a = b = c \Leftrightarrow \Delta ABC$đều.

Theo công thức tính độ dài phân giác, và áp dụng BĐT quen thuộc dạng $\frac{{4XY}}{{{{\left( {X + Y} \right)}^2}}} \le 1$ , ta có:
        $\begin{array}{l}
{l_a}^2{l_b}^2{l_c}^2 = \frac{{4bc}}{{{{\left( {b + c} \right)}^2}}}p\left( {p - a} \right).\frac{{4ca}}{{{{\left( {c + a} \right)}^2}}}p\left( {p - b} \right).\frac{{4ab}}{{{{\left( {a + b} \right)}^2}}}p\left( {p - c} \right)\\
 \le p\left( {p - a} \right).p\left( {p - b} \right).p\left( {p - c} \right)\\
 = {p^2}.p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)\\
 = {p^2}.{S^2}
\end{array}$
Suy ra: ${l_a}{l_b}{l_c} \le p.S$
Đẳng thức xảy ra $ \Leftrightarrow \Delta ABC$ đều.
bài này nhin quen quen .hi –  tathuynga2601 20-06-12 10:46 AM
thax bạn nhá :x –  leonguyen58 20-06-12 08:47 AM
mình đã giải chi tiết, bạn tham khảo nhé –  hoàng anh thọ 20-06-12 08:32 AM

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