Question

Cho tam giác ABC có BC = a ; CA = b ; AB = c. Chứng minh rằng:

a) \(sin\dfrac{A}{2}\)≤\(\dfrac{a}{b+c}\)

b) \(\sin\dfrac{A}{2}.\sin\dfrac{B}{2}.\sin\dfrac{C}{2}\) ≤ \(\dfrac{1}{8}\)

Answer 1

a, Vẽ phân giác AD của góc BAC

Kẻ BH\(\perp\)AD tại H ; CK\(\perp AD\) tại K

Dễ thấy \(sin\widehat{A_1}=sin\widehat{A_2}=sin\dfrac{A}{2}=\dfrac{BH}{AB}=\dfrac{CK}{AC}=\dfrac{BH+CK}{AB+AC}\le\)\(\le\dfrac{BD+CD}{b+c}=\dfrac{a}{b+c}\)

b, Tượng tự \(sin\dfrac{B}{2}\le\dfrac{b}{a+c};sin\dfrac{C}{2}\le\dfrac{c}{a+b}\)

Mặt khác \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)

\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)