Cho tam giác ABC, cmr a) cos a/2 = căn của p(p-a)/bc $cos\frac{A}{2}=$ $\sqrt{\frac{p(p-a)}{bc}}$ b) R ≥2r

a)
$\begin{array}{l} \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\  \Leftrightarrow 2{\cos ^2}\dfrac{A}{2} - 1 = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\  \Leftrightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{b^2} + {c^2} + 2bc - {a^2}}}{{2bc}}\\  \Leftrightarrow 2{\cos ^2}\dfrac{A}{2} = \dfrac{{{{\left( {b + c} \right)}^2} - {a^2}}}{{2bc}}\\  \Leftrightarrow {\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {b + c - a} \right)\left( {b + c + a} \right)}}{{4bc}}\\  \Leftrightarrow {\cos ^2}\dfrac{A}{2} = \dfrac{{4\left( {\dfrac{{b + c + a}}{2} - a} \right)\left( {\dfrac{{b + c + a}}{2}} \right)}}{{4bc}}\\  \Leftrightarrow {\cos ^2}\dfrac{A}{2} = \dfrac{{\left( {p - a} \right)p}}{{4bc}}\\  \Rightarrow \cos \dfrac{A}{2} = \sqrt {\dfrac{{4p\left( {p - a} \right)}}{{4bc}}}  = \sqrt {\dfrac{{p\left( {p - a} \right)}}{{bc}}}  \end{array}$

b)

Giả sử bất đẳng thức là đúng

$\begin{array}{l} R = \dfrac{{abc}}{{4S}},S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \\ r = \dfrac{S}{p} = \dfrac{{2S}}{{a + b + c}}\\ R \ge 2r \Leftrightarrow \dfrac{{abc}}{{4S}} \ge \dfrac{{2.2S}}{{a + b + c}}\\  \Leftrightarrow abc\left( {a + b + c} \right) \ge 16{S^2}\\  \Leftrightarrow abc\left( {a + b + c} \right) \ge 16.p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)\\  \Leftrightarrow abc\left( {a + b + c} \right) \ge 16.\dfrac{{a + b + c}}{2}.\left( {\dfrac{{a + b + c}}{2} - a} \right)\left( {\dfrac{{a + b + c}}{2} - b} \right)\left( {\dfrac{{a + b + c}}{2} - c} \right)\\  \Leftrightarrow abc\left( {a + b + c} \right) \ge \left( {a + b + c} \right)\left( {b + c - a} \right)\left( {a + c - b} \right)\left( {a + b - c} \right)\\  \Leftrightarrow abc \ge \left( {b + c - a} \right)\left( {a + c - b} \right)\left( {a + b - c} \right)\left( * \right)\\ \sqrt {\left( {b + c - a} \right)\left( {a + c - b} \right)}  \le \dfrac{{b + c - a + a + c - b}}{2} = c\left( {AM - GM} \right)\\ TT:\sqrt {\left( {b + c - a} \right)\left( {a + b - c} \right)}  \le b\, , \sqrt {\left( {a + c - b} \right)\left( {b + a - c} \right)}  \le a \end{array}$

Vậy $(*)$ đúng. Đẳng thức được chứng minh. Dấu bằng xảy ra khi $a=b=c$