Ta có:
$sin^2\dfrac{A}{2} = \dfrac{1 - cosA}{2}$
Áp dụng định lý $cosin$, ta được:
$\dfrac{1-cosA}{2} = \dfrac{1 -\dfrac{b^2 + c^2 - a^2}{2bc}}{2} = \dfrac{2bc - b^2 - c^2 +a^2}{4bc} = \dfrac{a^2 - (b-c)^2}{4bc}$
Xét bất đẳng thức: $\dfrac{a^2 - (b-c)^2}{4bc} \leq \left(\dfrac{a}{b+c}\right)^2$
$\Leftrightarrow [a^2 - (b-c)^2](b + c)^2 \leq 4a^2bc$
$\Leftrightarrow a^2(b + c)^2 - (b-c)^2(b+c)^2 \leq 4a^bc$
$\Leftrightarrow (ab + ac)^2 - (b-c)^2(b+c)^2 \leq 4a^2bc$
$\Leftrightarrow a^2b^2 + 2a^2bc + a^2c^2 \leq 4a^2bc + (b-c)^2(b+c)^2$
$\Leftrightarrow (b-c)^2(b+c)^2 - (a^2b^2 - 2a^2bc + a^2c^2) \geq 0$
$\Leftrightarrow (b-c)^2(b+c)^2 - a^2(b - c)^2 \geq 0$ $(*)$
$\Leftrightarrow (b-c)^2[(b+c)^2 - a^2] \geq 0$ $(*)$
Xét $ΔABC$ luôn có:
$AB + AC > BC \Rightarrow (AB + AC)^2 > BC^2 \Rightarrow (b + c)^2 > a^2$
$\Rightarrow (*)$ đúng
Do đó $sin^2\dfrac{A}{2} \leq \left(\dfrac{a}{b+c}\right)^2$
$\Rightarrow sin\dfrac{A}{2} \leq \dfrac{a}{b+c}$
Chứng minh tương tự, ta được:
$sin\dfrac{B}{2} \leq \dfrac{b}{a+c}$
$sin\dfrac{C}{2} \leq \dfrac{c}{a +b}$
$\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2} \leq \dfrac{a}{b+c}\cdot\dfrac{b}{a+c}\cdot \dfrac{c}{a +b} = \dfrac{abc}{(a+b)(b+c)(c+a)}$
Ta có: $(a+b)(b+c)(c+a) \geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca} = 8abc$
$\Rightarrow \dfrac{abc}{(a+b)(b+c)(c+a)} \leq \dfrac{abc}{8abc} = \dfrac{1}{8}$
$\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2} \leq \dfrac{1}{8}$
